NCERT Solutions for Class 12 Maths Chapter 11 – Three Dimensional Geometry Miscellaneous Exercise
Page No 497:
Question 1:
Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, −1), (4, 3, −1).
Answer:
Let OA be the line joining the origin, O (0, 0, 0), and the point, A (2, 1, 1).
Also, let BC be the line joining the points, B (3, 5, −1) and C (4, 3, −1).
The direction ratios of OA are 2, 1, and 1 and of BC are (4 − 3) = 1, (3 − 5) = −2, and (−1 + 1) = 0
OA is perpendicular to BC, if a1a2 + b1b2 + c1c2 = 0
∴ a1a2 + b1b2 + c1c2 = 2 × 1 + 1 (−2) + 1 ×0 = 2 − 2 = 0
Thus, OA is perpendicular to BC.
Question 2:
If l1, m1, n1 and l2, m2, n2 are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are m1n2 − m2n1, n1l2 − n2l1, l1m2 − l2m1.
Answer:
It is given that l1, m1, n1 and l2, m2, n2 are the direction cosines of two mutually perpendicular lines. Therefore,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7212/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m1b627aa5.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7212/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m6267a1ba.gif)
Let l, m, n be the direction cosines of the line which is perpendicular to the line with direction cosines l1, m1, n1and l2, m2, n2.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7212/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m6a35d65d.gif)
l, m, n are the direction cosines of the line.
∴l2 + m2 + n2 = 1 … (5)
It is known that,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7212/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_15715a27.gif)
∴ ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7212/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_52680be3.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7212/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_52680be3.gif)
Substituting the values from equations (5) and (6) in equation (4), we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7212/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_4926b6a6.gif)
Thus, the direction cosines of the required line are ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7212/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_6e5c7ae.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7212/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_6e5c7ae.gif)
Page No 498:
Question 3:
Find the angle between the lines whose direction ratios are a, b, c and b − c,
c − a, a − b.
Answer:
The angle Q between the lines with direction cosines, a, b, c and b − c, c − a,
a − b, is given by,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7214/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m35039106.gif)
Thus, the angle between the lines is 90°.
Question 4:
Find the equation of a line parallel to x-axis and passing through the origin.
Answer:
The line parallel to x-axis and passing through the origin is x-axis itself.
Let A be a point on x-axis. Therefore, the coordinates of A are given by (a, 0, 0), where a ∈ R.
Direction ratios of OA are (a − 0) = a, 0, 0
The equation of OA is given by,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7215/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_68ec3145.gif)
Thus, the equation of line parallel to x-axis and passing through origin is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7215/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_6541431e.gif)
Question 5:
If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (−4, 3, −6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.
Answer:
The coordinates of A, B, C, and D are (1, 2, 3), (4, 5, 7), (−4, 3, −6), and
(2, 9, 2) respectively.
The direction ratios of AB are (4 − 1) = 3, (5 − 2) = 3, and (7 − 3) = 4
The direction ratios of CD are (2 −(− 4)) = 6, (9 − 3) = 6, and (2 −(−6)) = 8
It can be seen that, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7218/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m5aa20779.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7218/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m5aa20779.gif)
Therefore, AB is parallel to CD.
Thus, the angle between AB and CD is either 0° or 180°.
Question 6:
If the lines
and
are perpendicular, find the value of k.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7219/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m3e1a5e16.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7219/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_14fd3bc6.gif)
Answer:
The direction of ratios of the lines,
and
, are −3, 2k, 2 and 3k, 1, −5 respectively.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7219/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m3e1a5e16.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7219/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_14fd3bc6.gif)
It is known that two lines with direction ratios, a1, b1, c1 and a2, b2, c2, are perpendicular, if a1a2 + b1b2 + c1c2 = 0
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7219/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m6e070bbc.gif)
Therefore, for
, the given lines are perpendicular to each other.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7219/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m1995e0e6.gif)
Question 7:
Find the vector equation of the plane passing through (1, 2, 3) and perpendicular to the plane![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7221/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m231c51c1.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7221/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m231c51c1.gif)
Answer:
The position vector of the point (1, 2, 3) is ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7221/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_18253dd8.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7221/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_18253dd8.gif)
The direction ratios of the normal to the plane,
, are 1, 2, and −5 and the normal vector is ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7221/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m427ecb5c.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7221/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_727d1ad7.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7221/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m427ecb5c.gif)
The equation of a line passing through a point and perpendicular to the given plane is given by, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7221/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m57de6e8a.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7221/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m57de6e8a.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7221/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_30fc5f80.gif)
Question 8:
Find the equation of the plane passing through (a, b, c) and parallel to the plane ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7223/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_bf57199.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7223/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_bf57199.gif)
Answer:
Any plane parallel to the plane,
, is of the form ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7223/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_ba884ca.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7223/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m76fb131.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7223/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_ba884ca.gif)
The plane passes through the point (a, b, c). Therefore, the position vector
of this point is ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7223/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m5091bb46.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7223/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_41cbb03f.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7223/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m5091bb46.gif)
Therefore, equation (1) becomes
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7223/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m21f5060.gif)
Substituting
in equation (1), we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7223/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m30b53617.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7223/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_2c55aa04.gif)
This is the vector equation of the required plane.
Substituting
in equation (2), we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7223/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m49cc2505.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7223/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m45589896.gif)
Question 9:
Find the shortest distance between lines ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7224/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m7ab0147e.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7224/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m7ab0147e.gif)
and
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7224/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m7c19837b.gif)
Answer:
The given lines are
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7224/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m7bc4e164.gif)
It is known that the shortest distance between two lines,
and
, is given by
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7224/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_31fb09d7.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7224/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m1ff5a798.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7224/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_45b48ead.gif)
Comparing
to equations (1) and (2), we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7224/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_70cc456f.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7224/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m284bf617.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7224/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m751c76a8.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7224/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_78ff4532.gif)
Substituting all the values in equation (1), we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7224/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_3143ee39.gif)
Therefore, the shortest distance between the two given lines is 9 units.
Question 10:
Find the coordinates of the point where the line through (5, 1, 6) and
(3, 4, 1) crosses the YZ-plane
Answer:
It is known that the equation of the line passing through the points, (x1, y1, z1) and (x2, y2, z2), is ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7226/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m5ad623c7.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7226/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m5ad623c7.gif)
The line passing through the points, (5, 1, 6) and (3, 4, 1), is given by,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7226/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m56472d8b.gif)
Any point on the line is of the form (5 − 2k, 3k + 1, 6 −5k).
The equation of YZ-plane is x = 0
Since the line passes through YZ-plane,
5 − 2k = 0
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7226/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m54f49232.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7226/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m35323d85.gif)
Therefore, the required point is
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7226/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m5aef6ecf.gif)
Question 11:
Find the coordinates of the point where the line through (5, 1, 6) and
(3, 4, 1) crosses the ZX − plane.
Answer:
It is known that the equation of the line passing through the points, (x1, y1, z1) and (x2, y2, z2), is ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7227/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m5ad623c7.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7227/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m5ad623c7.gif)
The line passing through the points, (5, 1, 6) and (3, 4, 1), is given by,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7227/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m56472d8b.gif)
Any point on the line is of the form (5 − 2k, 3k + 1, 6 −5k).
Since the line passes through ZX-plane,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7227/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m20746709.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7227/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_1642764f.gif)
Therefore, the required point is
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7227/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m737a8782.gif)
Question 12:
Find the coordinates of the point where the line through (3, −4, −5) and (2, − 3, 1) crosses the plane 2x + y + z = 7).
Answer:
It is known that the equation of the line through the points, (x1, y1, z1) and (x2, y2, z2), is ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7229/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m5ad623c7.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7229/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m5ad623c7.gif)
Since the line passes through the points, (3, −4, −5) and (2, −3, 1), its equation is given by,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7229/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m20f753da.gif)
Therefore, any point on the line is of the form (3 − k, k − 4, 6k − 5).
This point lies on the plane, 2x + y + z = 7
∴ 2 (3 − k) + (k − 4) + (6k − 5) = 7
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7229/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m5bdc5308.gif)
Hence, the coordinates of the required point are (3 − 2, 2 − 4, 6 × 2 − 5) i.e.,
(1, −2, 7).
Question 13:
Find the equation of the plane passing through the point (−1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.
Answer:
The equation of the plane passing through the point (−1, 3, 2) is
a (x + 1) + b (y − 3) + c (z − 2) = 0 … (1)
where, a, b, c are the direction ratios of normal to the plane.
It is known that two planes,
and
, are perpendicular, if ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7230/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_6632f0e4.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7230/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_202997ed.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7230/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_42b4865d.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7230/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_6632f0e4.gif)
Plane (1) is perpendicular to the plane, x + 2y + 3z = 5
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7230/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_1f844f9d.gif)
Also, plane (1) is perpendicular to the plane, 3x + 3y + z = 0
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7230/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m126f39d6.gif)
From equations (2) and (3), we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7230/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m35b42b.gif)
Substituting the values of a, b, and c in equation (1), we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7230/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_33a471cb.gif)
This is the required equation of the plane.
Question 14:
If the points (1, 1, p) and (−3, 0, 1) be equidistant from the plane
, then find the value of p.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7232/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_49db98db.gif)
Answer:
The position vector through the point (1, 1, p) is ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7232/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_35b50b34.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7232/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_35b50b34.gif)
Similarly, the position vector through the point (−3, 0, 1) is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7232/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_2447c6bc.gif)
The equation of the given plane is ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7232/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m18bad3cd.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7232/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m18bad3cd.gif)
It is known that the perpendicular distance between a point whose position vector is
and the plane,
is given by, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7232/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_507ffa15.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7232/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_4f2bfef5.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7232/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_5281f59e.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7232/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_507ffa15.gif)
Here,
and d![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7232/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_20f3b06c.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7232/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_20a1a5ec.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7232/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_20f3b06c.gif)
Therefore, the distance between the point (1, 1, p) and the given plane is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7232/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_4dc12a30.gif)
Similarly, the distance between the point (−3, 0, 1) and the given plane is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7232/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m766ee33d.gif)
It is given that the distance between the required plane and the points, (1, 1, p) and (−3, 0, 1), is equal.
∴ D1 = D2
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7232/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_4de7ad94.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7232/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m5b259922.gif)
Question 15:
Find the equation of the plane passing through the line of intersection of the planes
and
and parallel to x-axis.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7234/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_11ffb182.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7234/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_mfe1e333.gif)
Answer:
The given planes are
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7234/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_2dbcc5ff.gif)
The equation of any plane passing through the line of intersection of these planes is
r→.i⏜+j⏜+k⏜-1+λr→.2i⏜+3j⏜-k⏜+4=0r→.2λ+1i⏜+3λ+1j⏜+1-λk⏜+4λ-1=0 …(1)
Its direction ratios are (2λ + 1), (3λ + 1), and (1 − λ).
The required plane is parallel to x-axis. Therefore, its normal is perpendicular to x-axis.
The direction ratios of x-axis are 1, 0, and 0.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7234/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m150de5cb.gif)
Substituting
in equation (1), we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7234/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_77495372.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7234/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_1555842f.gif)
Therefore, its Cartesian equation is y − 3z + 6 = 0
This is the equation of the required plane.
Question 16:
If O be the origin and the coordinates of P be (1, 2, −3), then find the equation of the plane passing through P and perpendicular to OP.
Answer:
The coordinates of the points, O and P, are (0, 0, 0) and (1, 2, −3) respectively.
Therefore, the direction ratios of OP are (1 − 0) = 1, (2 − 0) = 2, and (−3 − 0) = −3
It is known that the equation of the plane passing through the point (x1, y1 z1) is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7235/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m1a7aa7e2.gif)
Here, the direction ratios of normal are 1, 2, and −3 and the point P is (1, 2, −3).
Thus, the equation of the required plane is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7235/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m2d3dc127.gif)
Question 17:
Find the equation of the plane which contains the line of intersection of the planes
,
and which is perpendicular to the plane
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7236/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_68803b22.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7236/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_259fe139.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7236/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_632c75c5.gif)
Answer:
The equations of the given planes are
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7236/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_mfb93068.gif)
The equation of the plane passing through the line intersection of the plane given in equation (1) and equation (2) is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7236/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m45613b95.gif)
The plane in equation (3) is perpendicular to the plane, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7236/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m324d3ed3.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7236/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m324d3ed3.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7236/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_2ed534c4.gif)
Substituting
in equation (3), we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7236/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m7824ae62.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7236/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_md708ef5.gif)
This is the vector equation of the required plane.
The Cartesian equation of this plane can be obtained by substituting
in equation (3).
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7236/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m49cc2505.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7236/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_1fb082b1.gif)
Page No 499:
Question 18:
Find the distance of the point (−1, −5, −10) from the point of intersection of the line
and the plane
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7238/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m13721434.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7238/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_65564f94.gif)
Answer:
The equation of the given line is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7238/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_31108b30.gif)
The equation of the given plane is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7238/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m6327323f.gif)
Substituting the value of
from equation (1) in equation (2), we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7238/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_41cbb03f.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7238/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_45898fa0.gif)
Substituting this value in equation (1), we obtain the equation of the line as
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7238/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_137e9bfb.gif)
This means that the position vector of the point of intersection of the line and the plane is ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7238/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_137e9bfb.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7238/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_137e9bfb.gif)
This shows that the point of intersection of the given line and plane is given by the coordinates, (2, −1, 2). The point is (−1, −5, −10).
The distance d between the points, (2, −1, 2) and (−1, −5, −10), is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7238/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m3978a7ff.gif)
Question 19:
Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes
and
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7239/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_434c3ad0.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7239/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_b754f6c.gif)
Answer:
Let the required line be parallel to vector
given by,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7239/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_6d43734a.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7239/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m182a051d.gif)
The position vector of the point (1, 2, 3) is ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7239/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m1662cbf4.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7239/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m1662cbf4.gif)
The equation of line passing through (1, 2, 3) and parallel to
is given by,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7239/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_6d43734a.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7239/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_3e9a99d0.gif)
The equations of the given planes are
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7239/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m7b56d0fb.gif)
The line in equation (1) and plane in equation (2) are parallel. Therefore, the normal to the plane of equation (2) and the given line are perpendicular.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7239/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m1f5b406b.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7239/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_56a2581.gif)
From equations (4) and (5), we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7239/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_809c22d.gif)
Therefore, the direction ratios of
are −3, 5, and 4.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7239/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_6d43734a.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7239/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m10bc00b6.gif)
Substituting the value of
in equation (1), we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7239/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_6d43734a.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7239/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_481bfd24.gif)
This is the equation of the required line.
Question 20:
Find the vector equation of the line passing through the point (1, 2, − 4) and perpendicular to the two lines: ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7241/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_5508a997.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7241/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_5508a997.gif)
Answer:
Let the required line be parallel to the vector
given by, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7241/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m182a051d.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7241/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_6d43734a.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7241/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m182a051d.gif)
The position vector of the point (1, 2, − 4) is ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7241/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m1674439.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7241/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m1674439.gif)
The equation of the line passing through (1, 2, −4) and parallel to vector
is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7241/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_6d43734a.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7241/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m7e49e3a.gif)
The equations of the lines are
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7241/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_3b656542.gif)
Line (1) and line (2) are perpendicular to each other.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7241/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m7ba9c19.gif)
Also, line (1) and line (3) are perpendicular to each other.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7241/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m7f14cf8d.gif)
From equations (4) and (5), we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7241/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_78d9b900.gif)
∴Direction ratios of
are 2, 3, and 6.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7241/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_6d43734a.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7241/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_4fde91c0.gif)
Substituting
in equation (1), we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7241/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_17eb5807.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7241/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_33a7b614.gif)
This is the equation of the required line.
Question 21:
Prove that if a plane has the intercepts a, b, c and is at a distance of P units from the origin, then ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7242/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m4a73d372.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7242/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m4a73d372.gif)
Answer:
The equation of a plane having intercepts a, b, c with x, y, and z axes respectively is given by,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7242/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_3c9715d4.gif)
The distance (p) of the plane from the origin is given by,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7242/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m118ea3e0.gif)
Question 22:
Distance between the two planes:
and
is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7244/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m7de5dc2.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7244/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_7704122.gif)
(A)2 units (B)4 units (C)8 units
(D)![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7244/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_6fe4ece6.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7244/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_6fe4ece6.gif)
Answer:
The equations of the planes are
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7244/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_4bf52274.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7244/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_7704122.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7244/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m54888ccd.gif)
It can be seen that the given planes are parallel.
It is known that the distance between two parallel planes, ax + by + cz = d1 and ax + by + cz = d2, is given by,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7244/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_m32df7e89.gif)
Thus, the distance between the lines is
units.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7244/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_3d6fc686.gif)
Hence, the correct answer is D.
Question 23:
The planes: 2x − y + 4z = 5 and 5x − 2.5y + 10z = 6 are
(A) Perpendicular (B) Parallel (C) intersect y-axis
(C) passes through ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7245/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_73c07216.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7245/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_73c07216.gif)
Answer:
The equations of the planes are
2x − y + 4z = 5 … (1)
5x − 2.5y + 10z = 6 … (2)
It can be seen that,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7245/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_7969a96d.gif)
∴ ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7245/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_5b8553c6.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/240/7245/NS_14-11-08_Gopal_12_Math_Exercise%2011.1_5_MNK_SS_html_5b8553c6.gif)
Therefore, the given planes are parallel.
Hence, the correct answer is B.