ML Aggarwal Class 12 Maths Solutions Section B Chapter 2 Three Dimensional Geometry Ex 2.6

Well-structured ISC Mathematics Class 12 Solutions Chapter 2 Three Dimensional Geometry Ex 2.6 facilitate a deeper understanding of mathematical principles.

ML Aggarwal Class 12 Maths Solutions Section B Chapter 2 Three Dimensional Geometry Ex 2.6

Very short answer type questions (1 to 2) :

Question 1.
(i) Write the intercept cut off by the plane 2x +y – z 5 on x-axis.
(ii) Write the intercept cut off by the plane 3x – 4y + z = 6 on y-axis.
(iii) Write the intercept cut off by the plane 5x – 2y + 3z + 7 = 0 on z-axis.
Solution:
(i) eqn. of given plane be 2x +y – z = 5
i.e. \(\frac{x}{\frac{5}{2}}+\frac{y}{5}+\frac{z}{-\frac{1}{5}}\) = 1 ………………(i)
∴ intercept made by given plane on x-axis be \(\frac{5}{2}\).
[putting y = 0, z = 0 in eqn. (1)]

(ii) eqn. of given plane be 3x – 4y + z = 6
i.e. \(\frac{3 x}{6}-\frac{4 y}{6}+\frac{z}{6}\) = 1
⇒ \(\frac{x}{2}+\frac{y}{-\frac{3}{2}}+\frac{z}{6}\) = 1
∴ intercept made by plane on y-axis be – \(\frac{3}{2}\).
[x = 0 and z = 0 in given eqn. of plane]

(iii) eqn. of given plane be 5x – 2y + 3z = – 7
⇒ \(\frac{5 x}{-7}+\frac{2 y}{7}-\frac{3 z}{7}\) = 1
⇒ \(\frac{x}{-\frac{7}{5}}+\frac{y}{\frac{7}{2}}+\frac{z}{-\frac{7}{3}}\) = 1
Thus, the plane cut off by given plane on Z- axis = – \(\frac{7}{3}\) [putting x = 0 = y in given plane].

Question 2.
(i) Find the intercepts made by the plane 2x + y- z = 5 on the coordinate axes. (NCERT)
(ii) Write the sum of intercepts cut off by the plane \(\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})\) – 5 = 0 on the coordinate axes.
Solution:
(i) Eqn. of given plane be 2x + y – z = 5
⇒ \(\frac{2 x}{5}+\frac{y}{5}-\frac{z}{5}\) = 1
⇒ \(\frac{x}{\frac{5}{2}}+\frac{y}{5}+\frac{z}{-5}\) = 1
which is of the form \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\)= 1
Thus, the intercept cut off by plane on coordinate axes are \(\frac{5}{2}\), 5 and – 5.

(ii) Vector eqn. of plane be
\(\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})\) = 5
Its cartesian form be
\((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(2 \hat{i}+\hat{j}-\hat{k})\) = 5
⇒ 2x + y – z = 5
⇒ \(\frac{x}{\frac{5}{2}}+\frac{y}{5}+\frac{z}{-5}\) = 1
The eqn. of plane in intercept form be
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\) = 1
i.e. a = \(\frac{5}{2}\) ;
b = 5 ;
c = – 5
Thus, sum of intercepts = a + b + c
= \(\frac{5}{2}\) + 5 – 5
= \(\frac{5}{2}\)

Question 3.
(i) Write the equation of the plane which has intercepts 2, 3 and 4 on the x-axis, y- axis and z-axis respectively. (NCERT)
(ii) Write the equation of the plane with intercepts 2, – 3 and 5 on the x-axis, y- axis and z-axis respectively.
Solution:
(i) Let the equation of plane be
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\) = 1 ………………(1)
where a, b, c be the intercepts made by plane (1) with coordinate axes.
∴ given a = 2, b = 3 and c = 4
∴ eqn. (1) becomes ;
\(\frac{x}{2}+\frac{y}{3}+\frac{z}{4}\) = 1
i.e. 6x + 4y + 3z = 12
which is the required eqn. of plane.

(ii) Let the equation of plane be
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\) = 1 ………………..(1)
where a, b, c be the length of intercepts cut
off by plane (1) on coordinate axes.
given a = 2 ; b = – 3 and c = 5
∴. eqn.(1) becomes;
⇒ 15x – 10y + 6z = 30
which is the required eqn. of plane.

Question 4.
(i) Find the equation of the plane passing through the points (2, 3, – 4) and (1, – 1, 3) and parallel to x-axis.
(ii) Find the equation of the plane through the points (2, 2, – 1) and (3, 4, 2) and parallel to the line whose direction ratios are < 7, 0, 6 >.
Solution:
(i) equation of plane through the point (2, 3, – 4) is given by
a (x – 2) + b (y – 3) + c (z + 4) = 0 …………………(1)
Now eqn. (1) passes through (1, – 1, 3).
∴ a (1 – 2) + b (- 1 – 3) + c (3 + 4) = 0
i.e. – a – 4b + 7c = 0
a + 4b – 7c = 0 ………………………(2)
Since the plane (1) is || to x-axis
∴ normal to plane (1) is ⊥ to x-axis whose
direction numbers are < 1, 0, 0 >
i.e. a + 0b + 0c = 0 …………………(3)
On solving eqn. (2) and eqn. (3) ; we have
\(\frac{a}{0}=\frac{b}{-7}=\frac{c}{-4}\) = k (say)
∴ a = 0; b = – 7k; c = – 4k
∴ from (1) ; we have
0 (x – 2) – 7k (y – 3) – 4k (z + 4) = 0
⇒ – 7y – 4z + 5 = 0
⇒ 7y + 4z – 5 = 0
which is the required eqn. of plane.

(ii) Equation of any plane through the point (2, 2, – 1) isgiven by
a (x – 2) + b (y – 2) + c (z + 1) = 0 ……………..(1)
Since eqn. (1) passes through the point (3, 4, 2).
∴ a (3 – 2) + b (4 – 2) + c (2 + 1) = 0
i.e. a + 2b + 3c = 0 …………………(2)
Also plane (1) is parallel to line whose direction ratios are < 7, 0, 6 >
∴ normal to plane is ⊥ to the line with direction ratios < 7, 0, 6 >
∴ 7a + 0b + 6c = 0 ………………(3)
On solving eqn. (2) and (3); we have
\(\frac{a}{12}=\frac{b}{21-6}=\frac{c}{0-14}\)
i.e. \(\frac{a}{12}=\frac{b}{15}=\frac{c}{-14}\) = k (say)
∴ a = 12k ; b = 15k and c = 14k
∴ eqn. (1) becomes ;
12 (x – 2) + 15k (y – 2) -14k (z + 1) = 0
⇒ 12x + 15y – 14z – 68 = 0 which is the required eqn. of plane.

Question 5.
Find the equation of the plane which passes through the point (3, 2, 0) and contains the line \(\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}\).
Solution:
Eqn. of plane through the point (3, 2, 0) is given by
a (x – 3) + b (y – 2) + c (z – 0) = 0 ……………….(1)
Since the plane (1) contains the given line
\(\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}\) ………………….(2)
∴ plane (1) passes through (3, 6, 4) and normal to plane (1) is ⊥to line (2).
i.e. a (3 – 3) + b (6 – 2) + c (4 – 0) = 0
i.e. 0a + 4b + 4c = 0
⇒ 0a + b + c = 0 ………………(3)
and 1a + 5b + 4c = 0 …………………(4)
On solving eqn. (3) and (4) ; we have
\(\frac{a}{4-5}=\frac{b}{1-0}=\frac{c}{0-1}\)
i.e. \(\frac{a}{-1}=\frac{b}{1}=\frac{c}{-1}\) = k (say)
∴ a = – k ;b = k and c = – k
∴ eqn. (1) becomes;
– k (x – 3) + k (y – 2) – kz = 0
⇒ – x + y – z + 1 = 0
⇒ x – y + z – 1 = 0
which is the required eqn. of plane.

Question 6.
Find the foot of perpendicular drawn from the point p (1, 2, 3) on the line \(\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}\). Also obtain the equation of the plane containing the line and the point (1, 2, 3).
Solution:
The eqn. of any plane containing the line
\(\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}\) ………………….(*)
is given by
a (x – 6) + b (y – 7) + c (z – 7) = 0 …………………….(1)
where 3a + 2b – 2c = 0 ………………….(2)
Also point (1, 2, 3) lies on plane (1) ; we get
– 5a – 5b – 4c = 0 ……………………..(3)
Solving (2) and (3) ; we get
\(\frac{a}{-18}=\frac{b}{22}=\frac{c}{-5}\) = k (say) ;
where k ≠ 0
∴ a = – 18k ;
b = 22k
and c = – 5k,
putting all these values in (1) ; we get
– 18k (x – 6) + 22k (y – 7) – 5k (z – 7) = 0
⇒ -18x + 22y – 5z – 11 = 0
⇒ 18x – 22y + 5z + 11 = 0 ………………….(4)
is the required eqn. of plane.
Any point on line (*) is M (3t + 6, 2t + 7, – 2t + 7)
∴ D’ ratios of PM are < 3t + 6 – 1, 2t + 7 – 2, – 2t + 7 – 3 >
i.e. < 3t + 5, 2t + 5, – 2t + 4 >
Now line PM is ⊥ to the line given by (*)
∴ (3t + 5) 3 + (2t + 5) 2 – 2 (- 2t + 4) = 0
⇒ 17t + 17 = 0
⇒ t = – 1
∴ Coordinates of foot of ⊥ is M (3, 5, 9).

Question 7.
(i) Find the equation of the plane passing through the points (- 1, 2, 0), (2, 2, – 1) and parallel to the line \(\frac{x-1}{1}=\frac{2 y+1}{2}=\frac{z+1}{-1}\).
(ii) Find the vector equation of the plane passing through the points (0, – 4, 1) and (2, 3, – 2) and parallel to the line \(\vec{r}=5 \hat{i}-2 \hat{j}+\hat{k}+\lambda(2 \hat{i}-\hat{k})\).
Solution:
(i) The eqn. of any plane passing through the point (- 1, 2, 0) is given by
a (x + 1) + b (y – 2) + c (z – 0) = 0 …………………(1)
Since plane (1) passes through the point (2, 2, – 1).
∴ a (2 + 1) + b (2 – 2) + c (- 1 – 0) = 0
i.e. 3a + 0b – c = 0 ……………..(2)
also eqn. of given line be
\(\frac{x-1}{1}=\frac{y+1 / 2}{1}=\frac{z+1}{-1}\) …………………(3)
Since plane (1) is parallel to line (3).
∴ a + b – c = 0 ……………………(3)
Solving eqn. (2) and eqn. (3) by crossmultiplication we get
\(\frac{a}{0+1}=\frac{b}{-1+3}=\frac{c}{3-0}\)
i.e. \(\frac{a}{1}=\frac{b}{2}=\frac{c}{3}\) = k (say)
∴ a = k ;
b = 2k
and c = 3k
putting the values of a, b and e in eqn. (1); we have
k (x + 1) + 2k (y – 2) + 3kz = 0
⇒ x + 2y + 3z = 3 be the required eqn. of plane.

(ii) Eqn. of any plane through the point (0, – 4, 1) is given by
a (x – 0) + b (y + 4) + c (z – 1) = 0 ………………(1)
Since eqn. (1) passes through the point (2, 3, – 2).
∴ a (2 – 0) + b (3 + 4) + c (- 2 – 1) =0
i.e. 2a + 7b – 3c = 0 ……………………..(2)
Also eqn. of given line be,
\(\vec{r}=5 \hat{i}-2 \hat{j}+\hat{k}+\lambda(2 \hat{i}-\hat{k})\) …………………….(3)
∴ given line is || to vector \(2 \hat{i}-\hat{k}\)
Thus, D’ No’s of the line be < 2, 0, – 1 >
Since the plane is || to line (2)
∴ normal to plane is ⊥ to given line (3)
∴ 2a + 0b – c = 0 …………………….(4)
On solving (2) and (4) ; we have
\(\frac{a}{-7-0}=\frac{b}{-6+2}=\frac{c}{0-14}\)
i.e. \(\frac{a}{-7}=\frac{b}{-4}=\frac{c}{-14}\) = k (say)
⇒ a = – 7k ;
b = – 4k ;
c = – 14k
∴ eqn. (1) becomes ;
⇒ – 7kx – 4k (y + 4) – 14k (z – 1) = 0
⇒ – 7x – 4y – 14z – 2 = 0
⇒ 7x + 4y + 14z + 2 = 0 ……………………(2)
If \(\vec{r}\) be the P.V of point P (x, y, z).
Then \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\)
∴ eqn. (5) can be written as;
\((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(7 \hat{i}+4 \hat{j}+14 \hat{k})\) + 2 = 0
⇒ \(\vec{r} \cdot(7 \hat{i}+4 \hat{j}+14 \hat{k})\) + 2 = 0

Question 7 (old).
(i) Find the cartesian equation of the plane passing through the points A (0, 0, 0) and B (3, – 1, 2) and parallel to the line \(\frac{x-4}{1}=\frac{y+3}{-4}=\frac{z+1}{7}\).
Solution:
eqn. of any plane through (0, 0, 0) is
a(x – 0) + b (y – 0) + c (z – 0) = 0 …………………(1)
eqn. (1) passes through (3, – 1, 2) ; we have
3a – b + 2c = 0 ………………….(2)
Since the plane (1) is parallel to given line
\(\frac{x-4}{1}=\frac{y+3}{-4}=\frac{z+1}{7}\)
∴ a – 4b + 7c = 0 ……………………….(3)
On solving (2) and (3) ; we have
\(\frac{a}{1}=\frac{b}{-19}=\frac{c}{-11}\) = k (say) ; where k ≠ 0
∴ a = k ;
b = – 19k
and c = – 11k
putting all these values in (1) ; we get
kx – 19k – 11kz = 0
i.e. x – 19y – 11z = 0 is the required eqn. of plane.

Question 8.
Find the equation of the plane passing through the point (1, 2, – 4) and parallel to the lines \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+1}{6}\) and \(\frac{x-1}{1}=\frac{y+3}{1}=\frac{z}{-1}\).
Solution:
the required equation of plane through the point (1, 2, – 4) and parallel to given lines
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+1}{6}\)
and \(\frac{x-1}{1}=\frac{y+3}{1}=\frac{z}{-1}\)
i.e. parallel to lines whose direction numbers are < 2, 3, 6 > and < 1, 1, – 1 >
We know that, eqn. of plane passing through the point (x₁, y₁, z₁)
and || to lines having direction ratios < a₁, b₁, c₁ > and < a₂, b₂, c₂ > is
\(\left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|\) = 0
Here a₁ = 2 ;
b₁ = 3 ;
c₁ = 6 ;
a₂ = 1 ;
b₂ = 1,
c₂ = – 1
∴ required eqn. of plane be
\(\left|\begin{array}{ccc}
x-1 & y-2 & z+4 \\
2 & 3 & 6 \\
1 & 1 & -1
\end{array}\right|\) = 0
⇒ (x – 1) (- 3 – 6) – (y – 2) (- 2 – 6) + (z + 4) (2 – 3) = 0
⇒ – 9x + 9 + 8y – 16 – z – 4 = 0
⇒ – 9x + 8y – z – 11 = 0
⇒ 9x – 8y + z + 11 = 0
which is the required eqn. of plane.

Question 8 (old).
(i) Find the equation of the plane passing through origin and which is parallel to the vectors \(\hat{i}+\hat{j}-\hat{k}\) and \(3 \hat{i}-\hat{k}\).
Solution:
The required eqn. of plane passing through (0, 0, 0) and || to vectors \(\hat{i}+\hat{j}-\hat{k}\) and \(3 \hat{i}-\hat{k}\)
i.e. to lines whose direction numbers are < 1, 1, – 1 > and < 3, 0, – 1 >
We know that eqn. of plane through (x₁, y₁, z₁)
and || to lines with direction numbers are < a₁, b₁, c₁ > and < a₂, b₂, c₂ > is
\(\left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|\) = 0
∴ required eqn. of plane be
\(\left|\begin{array}{ccc}
x-0 & y-0 & z-0 \\
1 & 1 & -1 \\
3 & 0 & -1
\end{array}\right|\) = 0
⇒ x (- 1 – 0) – y (- 1 + 3) + z (0 – 3) = 0
⇒ – x – 2y – 3z = 0
⇒ x + 2y + 3z = 0
be the required eqn. of plane.

Question 9.
Find the vector as well as the cartesian equation of the plane passing through the point (1, 2, – 4) and parallel to the lines \(\vec{r}=\hat{i}+2 \hat{j}-\hat{k}+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})\) and \(\vec{r}=\hat{i}-3 \hat{j}+5 \hat{k}+\mu(\hat{i}+\hat{j}-\hat{k})\).
Solution:
The required equation of plane through the point (1, 2, – 4) and parallel to given lines
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+1}{6}\)
and \(\frac{x-1}{1}=\frac{y+3}{1}=\frac{z}{-1}\)
i.e. parallel to lines whose direction numbers are < 2, 3, 6 > and < 1, 1, – 1 >
We know that, eqn. of plane passing through the point (x₁, y₁, z₁)
and || to lines having direction ratios < a₁, b₁, c₁ > and < a₂, b₂, c₂ > is
\(\left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|\) = 0
Here a₁ = 2 ;
b₁ = 3 ;
c₁ = 6 ;
a₂ = 1
b₂ = 1
c₂ = – 1
∴ required eqn. of plane be
\(\left|\begin{array}{ccc}
x-1 & y-2 & z+4 \\
2 & 3 & 6 \\
1 & 1 & -1
\end{array}\right|\) = 0
⇒ (x – 1) (- 3 – 6) – (y – 2) (- 2 – 6) + (z + 4) (2 – 3) = 0
⇒ – 9x + 9 + 8y – 16 – z – 4 = 0
⇒ – 9x + 8y – z – 11 = 0
⇒ 9x – 8y + z + 11 = 0
which is the required eqn. of plane in cartesian form
Also its vector form be
\(\vec{r} \cdot(9 \hat{i}-8 \hat{j}+\hat{k})\) + 11 = 0
[where \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\) be the position vector of any point P(x, y, z) on plane]

Question 10.
Find the equation of the plane containing the line \(\frac{x-4}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) and parallel to the line \(\frac{x}{1}=\frac{y-1}{3}=\frac{z-2}{3}\).
Solution:
Given eqns. of lines are ;
\(\frac{x-4}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) ……………..(1)
and \(\frac{x}{1}=\frac{y-1}{3}=\frac{z-2}{3}\) ………………………(2)
eqn. of plane containing eqn. (1) is given by
a (x – 4) + b (y – 2) + c (z – 3) = 0 …………….(3)
and 2a + 3b + 4c = 0 ………………..(4)
Also plane (1) parallel to line (2)
∴ Normal to plane is ⊥ to the line (2).
∴ a + 3b + 3c = 0 ………………….(5)
On solving eqn. (4) and eqn. (5) ; we have
\(\)
i.e. \(\) = k (say)
∴ a = – 3k ;
b = – 2k ;
c = 3k
Thus, eqn. (3) becomes ;
⇒ – 3k (x – 4) – 2k (y- 2) + 3k (z – 3) = 0
⇒ – 3x – 2y + 3z + 7 = 0
= 3x + 2y – 3z – 7 = 0
which is the required eqn. of plane.

Question 11.
Find the vector equation of the plane which contains the two parallel lines \(\frac{x-4}{1}=\frac{y-3}{-4}=\frac{z-2}{5}\) and \(\frac{x-3}{1}=\frac{y+2}{-4}=\frac{z}{5}\).
Solution:
equation of given lines are
\(\frac{x-4}{1}=\frac{y-3}{-4}=\frac{z-2}{5}\) …………………..(1)
and \(\frac{x-3}{1}=\frac{y+2}{-4}=\frac{z}{5}\) …………………(2)
equation of any plane containing line (1) is
given by a (x – 4) + b (y – 3) + c(z – 2) = 0 ……………..(3)
and a – 4b + 5c = 0 ………………(4)
Also line (2) lies in plane (3).
∴ the point (3, – 2, 0) lies on plane (3) ; we have
a (3 – 4) + b (- 2 – 3) + c (0 – 2) = 0
i.e. – a – 5b – 2c = 0 ……………….(5)
On solving eqn. (4) and (5) ; we have
\(\frac{a}{8+25}=\frac{b}{-5+2}=\frac{c}{-5-4}\)
i.e. \(\frac{a}{33}=\frac{b}{-3}=\frac{c}{-9}\)
i.e. \(\frac{a}{11}=\frac{b}{-1}=\frac{c}{-3}\) = k (say)
⇒ a = 11k ;
b = – k ;
c = – 3k ;
From (1) ;
⇒ 11k (x – 4) – k (y – 3) – 3k (z – 2) = 0
⇒ 11x – y – 3z = 35 be the required eqn. of plane in cartesian form.
Thus, in vector form, eqn. of plane be
\(\vec{r} \cdot(11 \hat{i}-\hat{j}-3 \hat{k})\) = 35
where \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\) be the P.V, of any point on plane.

Question 12.
If the lines \(\frac{x-1}{k}=\frac{y-2}{2}=\frac{z-3}{3}\) and \(\frac{x-2}{3}=\frac{y-3}{k}=\frac{z-1}{2}\) intersect at a point, find the value(s) of k.
Solution:
The given lines are,
\(\frac{x-1}{k}=\frac{y-2}{2}=\frac{z-3}{3}\) ……………………(1)
and \(\frac{x-2}{3}=\frac{y-3}{k}=\frac{z-1}{2}\) …………………..(2)
On comparing eqn. (1) and eqn. (2) with lines
\(\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}\) …………………….(3)
\(\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}\) ……………………(4)
Now lines (3) and (4) coplanar iff
⇒ \(\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|\) = 0
Since lines (1) and (2) are coplanar.
∴ \(\left|\begin{array}{ccc}
2-1 & 3-2 & 1-3 \\
k & 2 & 3 \\
3 & k & 2
\end{array}\right|\) = 0
⇒ \(\left|\begin{array}{ccc}
1 & 1 & -2 \\
k & 2 & 3 \\
3 & k & 2
\end{array}\right|\) = 0 ;
Excluding along R₁
1 (4 – 3k) – 1 (2k – 9) – 2(k² – 6) = 0
⇒ 4 – 3k – 2k + 9 – 2k² + 12 = 0
⇒ 2k² + 5k – 25 = 0
⇒ 2k² + 10k – 5k – 25 = 0
⇒ 2k (k + 5) – 5 (k + 5) = 0
⇒ (2k – 5) (k + 5) = 0
⇒ k = \(\frac{5}{2}\), – 5

Question 12 (old).
(i) If the lines \(\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}\) and \(\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}\) are coplanar, find the value(s) of k.
Solution:
The given lines are,
\(\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}\) ………………..(1)
and \(\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}\) …………………(2)
On comparingeqn. (1) and eqn. (2) with lines
\(\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}\) …………………(3)
\(\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}\) …………………(4)
Now lines (3) and (4) coplanar iff
\(\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|\) = 0
Since lines (1) and (2) coplanar.

∴ \(\left|\begin{array}{ccc}
1-2 & 4-3 & 5-4 \\
1 & 1 & -k \\
k & 2 & 1
\end{array}\right|\) = 0
⇒ \(\left|\begin{array}{ccc}
-1 & 1 & 1 \\
1 & 1 & -k \\
k & 2 & 1
\end{array}\right|\) = 0 ;
Excluding along R₁
– 1 (1 + 2k) – 1(1 + k²) + 1 (2 – k) = 0
⇒ – 1 – 2k – 1 – k² + 2 – k = 0
⇒ – 3k – k² = 0
⇒ – k (3 + k) = 0
⇒ k = 0, – 3.

Question 13.
Show that the lines \(\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}\) and \(\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}\) are coplanar. Also find the equation of the plane containing the lines.
Solution:
We know that,
\(\frac{x-x_1}{l_1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1}\)
and \(\frac{x-x_2}{l_2}=\frac{y-y_2}{m_2}=\frac{z-z_2}{n_2}\) are coplanar
if \(\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
l_1 & m_1 & n_1 \\
l_2 & m_2 & n_2
\end{array}\right|\) = 0
and equation of plane containing given lines be given by
\(\left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
l_1 & m_1 & n_1 \\
l_2 & m_2 & n_2
\end{array}\right|\) = 0
Here x₁ = 3 ;
y₁ = 1 ;
z₁ = 5
x₂ = – 1
y₂ = 2
z₂ = 5
l₁ = – 3 ;
m₁ = 1 ;
n₁ = 5 ;
l₂ = – 1
m₂ = 2 ;
n₂ = 5
Here
\(\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
l_1 & m_1 & n_1 \\
l_2 & m_2 & n_2
\end{array}\right|\) = \(\left|\begin{array}{rrr}
2 & 1 & 0 \\
-3 & 1 & 5 \\
-1 & 2 & 5
\end{array}\right|\) ;
operate R₁ → R₁ + R₂
= \(\left|\begin{array}{ccc}
-1 & 2 & 5 \\
-3 & 1 & 5 \\
-1 & 2 & 5
\end{array}\right|\)
= 0
[∵ R₁ and R₃ are identical]
Thus given lines are coplanar.
and eqn. of plane containing the given lines is
given by \(\left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
l_1 & m_1 & n_1 \\
l_2 & m_2 & n_2
\end{array}\right|\) = 0
i.e. \(\left|\begin{array}{ccc}
x+3 & y-1 & z-5 \\
-3 & 1 & 5 \\
-1 & 2 & 5
\end{array}\right|\) = 0;
expanding along R₁
(x + 3) (- 5) – (y – 1) (- 10) + (z – 5) (- 5) = 0
⇒ – 5[x + 3 – 2y + 2 + z – 5] = 0
⇒ x – 2y + z = 0.

Question 14.
Show that the lines \(\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}\) and \(\frac{x}{1}=\frac{y-7}{-3}=\frac{z+7}{2}\) intersect. Find the point of intersection and the equation of the plane containing them.
Solution:
Given eqn’s of lines are ;
\(\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}\) …………………..(1)
and \(\frac{x}{1}=\frac{y-7}{-3}=\frac{z+7}{2}\) ………………….(2)
We know that,
\(\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}\)
and \(\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}\) intersects
if \(\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|\) = 0
Here,
x₁ = – 1 ;
y₁ = 3 ;
z₁ = – 2 ;
x₂ = 0 ;
y₂ = 7 ;
z₂ = – 7 ;
a₁ = – 3 ;
b₁ = 2 ;
c₁ = 1 ;
a₂ = 1;
b₂ = – 3 ;
c₂ = 2
Here,
\(\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|\)
= \(\left|\begin{array}{rrr}
1 & 4 & -5 \\
-3 & 2 & 1 \\
1 & -3 & 2
\end{array}\right|\) ;
Expanding along R₁
= 1 (4 + 3) – 4 (- 6 – 1) – 5 (9 – 2)
= 7 + 28 – 35 = 0
∴ Both lines given by eqn. (1) and eqn. (2) intersects
Any point on Iine(1) be P(- 3t – 1, 2t + 3, t – 2)
as the lines (1) and (2) intersects
∴ point P lies on eqn. (2) for some values of r.
\(\frac{-3 t-1}{1}=\frac{2 t+3-7}{-3}=\frac{t-2+7}{2}\)
i.e. \(\frac{-3 t-1}{1}=\frac{2 t-4}{-3}=\frac{t+5}{2}\)
from (1) and (2) fraction, we have
– 3t – 1 = \(\frac{2 t-4}{-3}\)
⇒ – 3 (- 3t – 1) = 2t – 4
⇒ 9t + 3 = 2t – 4
⇒ 7t = – 7
⇒ t = – 1
From (2) and (3) fraction; we have
4t – 8 = – 3t – 15
⇒ 7t = – 7
⇒ t = – 1
∴ required point of intersection of lines (1) and (2) be given by
P (3 – 1, – 2 + 3, – 1 – 2)
i.e. P(2, 1,- 3)
The required eqn. of plane be given by
\(\left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|\) = 0
i.e. \(\left|\begin{array}{ccc}
x+1 & y-3 & z+2 \\
-3 & 2 & 1 \\
1 & -3 & 2
\end{array}\right|\) = 0
⇒ (x + 1) (4 + 3) – (y – 3) (- 6 – 1) + (z + 2) (9 – 2) = 0
⇒ (x + 1) + 7 (y – 3) + 7 (z + 2) = 0
⇒ x +y + z = 0, which is the required eqn. of plane.

Question 15.
If the lines \(\frac{x-1}{-3}=\frac{y-2}{-2 k}=\frac{z-3}{1}\) and \(\frac{x-1}{k}=\frac{y-2}{1}=\frac{z-3}{5}\) are perpendicular, find the value of k and hence find the equation of the plane containing these lines.
Solution:
Eqns. of given lines are
\(\frac{x-1}{-3}=\frac{y-2}{-2 k}=\frac{z-3}{2}\) ………………….(1)
and \(\frac{x-1}{k}=\frac{y-2}{1}=\frac{z-3}{5}\) ……………….(2)
Since lines (1) and (2) are ⊥.
∴ (- 3) k – 2k (1) + 2 (5) = 0
[∵ a₁a₂ + b₁b₂ + c₁c₂ = 0]
⇒ – 5k + 10 = 0
⇒ k = 2
required eqn. of plane containing given lines be given by
\(\left|\begin{array}{crc}
x-1 & y-2 & z-3 \\
-3 & -4 & 2 \\
2 & 1 & 5
\end{array}\right|\) = 0 ;
expanding along R₁
⇒ (x – 1) (- 20 – 2) – (y – 2) (- 15 – 4) + (z – 3) (- 3 + 8) = O
⇒ (x – 1) (- 22) – (y – 2) (- 19) + (z – 3) 5 = 0
⇒ 22x – 19y – 5z + 31 = 0 be the required plane.

Question 16.
Find the equations of the planes that passes through the points :
(i) (1, 1, – 1), (6, 4, – 5) and (- 4, – 2, 1) (NCERT)
(ii) (2, 1,0), (3,-2,-2) and (3, 1, 7) (NCERT Exemplar)
(iii) A (2, 1, – 3), B (- 3, – 2, 1) and C (2, 4, – 1). (ISC 2011)
Solution:
(I) eqn. of plane through (1, 1, – 1) is given by
a (x – 1) + b (y – 1) + c (z + 1) = 0 …………………(1)
where < a, b, c > are the direction ratios of normal to plane.
Since plane (1) passes through the points (6, 4, – 5) and (- 4, – 2, 1)
∴ 5a + 3b – 4c = 0 …………………(2)
and – 5a – 3b + 2c = 0 ……………….(3)
On solving eqn. (2) and eqn. (3) ; we have
\(\frac{a}{6-12}=\frac{b}{20-10}=\frac{c}{-15+15}\)
i.e. \(\frac{a}{-6}=\frac{b}{10}=\frac{c}{0}\) = k ≠ 0
∴ a = – 6k ;
b = 10 k ;
c = 0
∴ from (1) ;
– 6k (x – 1) + 10k (y – 1) = 0
⇒ – 6x + 10y – 4 = 0
i.e. 3x – 5y + 2 = O which is the reqd. eqn. of plane.

(ii) eqn. of any plane through te point (2, 1, 0) is given by
a (x – 2) + b (y – 1) + c (z – 0) = 0 ……………….(1)
Since the plane passes through given points (3, – 2, – 2) and (3, 1, 7).
Thus (3, – 2, – 2) and (3, 1, 7) lies on eqn. (1).
a (3 – 2) + b (- 2 – 1) + c (- 2 – 0) = 0
⇒ a – 3b – 2c = 0 …………………(2)
also, a (3 – 2) + b (1 – 1) + c (7 – 0) = 0
⇒ a + 0b + 7c = 0 ………………..(3)
On solving (2) and (3) ; we have
\(\frac{a}{-21-0}=\frac{b}{-2-7}=\frac{c}{0+3}\)
i.e. \(\frac{a}{-21}=\frac{b}{-9}=\frac{c}{3}\) = k (say)
From eqn. (1) ; we have
– 21k (x – 2) – 9k (y – 1) + 3k (z – 0) = 0
⇒ – 3k [7x – 14 + 3y – 3 – z) = 0
⇒ 7x + 3y – z = 17 be the required equation of plane.

(iii) equation of plane passing through the point A (2, 1, – 3) is given by
a (x – 2) + b (y – 1) + c (z + 3) = 0 ………………….(1)
Since eqn. (1) passes through two given points
i.e. B (- 3, – 2, 1) and C (2, 4, – 1).
∴ a (- 3 – 2) + b (- 2 – 1) + c (1 + 3) = 0
i.e. – 5b – 3b + 4c = 0 ………………………(2)
and a (2 – 2) + b (4 – 1) + c (- 1 + 3) = 0
i.e. 0a + 3b + 2c = 0 ………………..(3)
On solving eqn. (2) and eqn. (3) ; we have
\(\frac{a}{-6-12}=\frac{b}{0+10}=\frac{c}{-15}\)
i.e. \(\frac{a}{-18}=\frac{b}{10}=\frac{c}{-15}\) = k (say)
∴ a = – 18k ;
b = 10k ;
c = – 15k
Thus, from eqn. (1); we have
– 18k (x – 2) + 10k (y – 1) – 15k (z + 3) = 0
⇒ – 18x + 10y – 15z – 19 = 0
⇒ 18x – 10y + 15z + 19 = 0
which is the required eqn. of plane.

Question 17.
Find the vector equation of the plane passing through points with position vectors \(2 \hat{i}+5 \hat{j}-3 \hat{k}\), \(-2 \hat{i}-3 \hat{j}+5 \hat{k}\) and \(5 \hat{i}+3 \hat{j}-3 \hat{k}\). (NCERT)
Solution:
The required plane passes through P (2, 5, – 3) having P.V.
\(\vec{a}=2 \hat{i}+5 \hat{j}-3 \hat{k}\) and is normal to the vector \(\vec{n}\) given by
\(\vec{n}=\overrightarrow{\mathrm{PQ}} \times \overrightarrow{\mathrm{PR}}\)

Question 17 (old).
Find the cartesian and the vector equations of the plane passing through the points R (2, 5, – 3), S (- 2, – 3, 5) and T (5, 3, – 3).
Solution:
The required plane passes through P (2, 5, – 3) having P.V.
\(\vec{a}=2 \hat{i}+5 \hat{j}-3 \hat{k}\) and is normal to the vector \(\vec{n}\) given by
\(\vec{n}=\overrightarrow{\mathrm{PQ}} \times \overrightarrow{\mathrm{PR}}\)

Question 18.
Show that the points (1, – 1, 1), (2, 3, 1), (1, 2, 3) and (0, – 2, 3) are coplanar. Also find the equation of the plane in which they lie.
Solution:
equation of plane through the point (1, – 1, 1) is given by
a (x – 1) + b (y + 1) + c (z – 1) = 0 ………………….(1)
Now plane (1) passes through the point (2, 3, 1).
∴ a (2 – 1) + b (3 + 1) + c (1 – 1) = 0
i.e. a + 4b + 0c = 0 …………….(2)
Also plane (1) passes through the point (1, 2, 3).
∴ a(1 – 1) + b (2 + 1) + c (3 – 1) = 0
i.e. 0a + 3b + 2c = 0 ………………….(3)
On solving eqn. (2) and eqn. (3) ; we have
\(\frac{a}{8-0}=\frac{b}{0-2}=\frac{c}{3-0}\)
i.e. \(\frac{a}{8}=\frac{b}{-2}=\frac{c}{3}\) = k (say)
i.e. a = 8k ;
b = – 2k ;
c = 3k
∴ eqn. (1) becomes ;
8k (x – 1) – 2k (y + 1) + 3k (z – 1) = 0
⇒ 8x – 2y + 3z – 13 = 0 ……………………(4)
Now the point (0, – 2, 3) lies on eqn. (4)
if 8 × 0 – 2 × (- 2) + 3 (3) – 13 = 0
if 0 = 0, which is true.
Hence the given four points lies on plane (4)
Thus given points are coplanar.

Question 19.
(i) Find the equations of the line passing through (2, – 1, 2) and (5, 3, 4) and of the plane passing through (2, 0, 3), (1, 1, 5) and (3, 2, 4). Also, find their point of intersection.
(ii) Find k such that the four points A (4, 4, 4), B (5, k, 8), C (5, 4, 1) and D (7, 7, 2) are coplanar.
Solution:
(i) The eqn. of line through two given points (2, – 1, 2) and (5, 3, 4) be given by
\(\frac{x-2}{5-2}=\frac{y+1}{3+1}=\frac{z-2}{4-2}\)
i.e. \(\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}\) ……………….(1)
Let the eqn. of plane through the point A (2, 0, 3) be given by
a (x – 2) + b (y – 0) + c (z – 3) = 0 ……………….(2)
The points B (1, 1, 5) and C (3, 2, 4) lies on plane (2)
∴ a (1 – 2) + b (1 – 0) + c (5 – 3) = 0
⇒ – a + b + 2c = 0 ……………..(3)
and a (3 – 2) + b (2 – 0) + c (4 – 3) = 0
⇒ a + 2b + c = 0 ……………….(4)
On solving eqn. (3) and (4); we have
\(\frac{a}{1-4}=\frac{b}{2+1}=\frac{c}{-2-1}\)
⇒ \(\frac{a}{-3}=\frac{b}{3}=\frac{c}{-3}\)
⇒ \(\frac{a}{1}=\frac{b}{-1}=\frac{c}{1}\)
Thus eqn. (2) becomes ;
1 (x – 2) – 1 (y – 0) + 1 (z – 3) = 0
⇒ x – y + z = 5 ………………….(5)
which is the required eqn. of plane.
Any point on line (1) be
P (3t + 2, 4t – 1, 2t + 2) ………………..(6)
Now point P be the point of intersection of line (1) and plane (5) 1ff point Plies on plane (5).
∴ 3t + 2 – (4t – 1) + 2t + 2 = 5
⇒ t + 5 = 5
⇒ t = 0
∴ from (6) ;
The coordinates of point of intersection are (3 × 0 + 2, 4 × 0 – 1, 2 × 0 + 2) i.e. (2, – 1, 2).

(ii) Eqn. of any plane through the point A (4, 4, 4) be given by
a(x – 4) + b (y – 4) + c(z – 4) = 0 ……………….(1)
where < a, b, c > are the direction numbers of normal to plane (1).
Now point C (5, 4, 1) lies on plane (1).
∴ a (5 – 4) + b (4 – 4) + c (1 – 4) = 0
⇒ a + 0b – 3c = 0
Also plane (1) pass through the point D (7, 7, 2)
∴ a (7 – 4) + b (7 – 4) + c (2 – 4) = 0
i.e. 3a + 3b – 2c = 0 ………………(3)
on solving eqn. (2) and eqn. (3) by cross-multiplication method, we have
9k (x – 4) – 7k (y – 4) + 3k (z – 4) = 0
⇒ 9x – 7y + 3z – 20 = 0 …………………..(4)
Since the given four points are coplanar so the fourth point B (5, k, 8) must lies on the plane through the remaining three points.
∴ 9 × 5 – 7 × k + 3 × 8 – 20 = 0
⇒ 45 – 7k + 24 – 20 = 0
⇒ 49 – 7k = 0
⇒ k = 7.

Question 20.
(i) Find the equation of the plane passing through the point (2, 4, 6) and making equal intercepts on axes.
(ii) Find the equation of the plane which passes through the point (1, – 2, – 1) and makes intercepts on the axes such that intercepts on y-axis and z-axis are respectively twice and thrice of the intercept made on x-axis.
Solution:
(i) Let the equations of plane having equal intercepts a on coordinate axes be
\(\frac{x}{a}+\frac{y}{a}+\frac{z}{a}\) = 1
i.e. x + y + z = a …………………..(1)
Since eqn. (1) passes through (2, 4, 6)
∴ 2 + 4 + 6 = a
⇒ a = 12
Thus from eqn. (1) ; we have
x + y ± z = 12 be the required eqn. of plane.

(ii) Let the equation of plane be
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\) = 1 …………………(1)
where a, b, c are the lengths of intercepts made by plane (1) on x-axis, y-axis and z-axis respectively.
Since plane (1) passes through (1, – 2, – 1).
∴ \(\frac{1}{a}-\frac{2}{b}-\frac{1}{c}\) = 1 ………………..(2)
It is given that, b = 2a and c = 3a
∴ eqn. (2) becomes ;
\(\frac{1}{a}-\frac{2}{2 a}-\frac{1}{3 a}\) = 1
⇒ – \(\frac{1}{3 a}\) = 1
⇒ a = – \(\frac{1}{3}\)
and b = – \(\frac{2}{3}\) ;
c = – 1
Thus eqn. (1) becomes ;
\(\frac{x}{-\frac{1}{3}}+\frac{y}{-\frac{2}{3}}+\frac{z}{-1}\) = 1
⇒ – 3x – \(\frac{3 y}{2}\) – z = 1
⇒ – 6x – 3y – 2z = 2
⇒ 6x + 3y + 2z + 2 = 0
which is the required eqn. of plane.

Question 20 (old).
Find the coordinates of the point where the line through (3, – 4, – 5) and (2, – 3, 1) crosses the plane determined by the points A (1, 2, 3), B (2, 2, 1) and C (- 1, 3, 6).
Solution:
Eqn. of line through the points (3, – 4, – 5) and (2, – 3, 1) is given by
\(\frac{x-3}{2-3}=\frac{y+4}{-3+4}=\frac{z+5}{1+5}\)
i.e. \(\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}\) …………….(1)
eqn. of plane through the point A (1, 2, 3) is given by
a(x – 1) + b (y – 2) + c (z – 3) = 0 ……………….(2)
Now, plane (2) passes through the point B (2, 2, 1).
∴ a (2 – 1) + b (2 – 2) + c (1 -3) = 0
a + 0b -2c = 0 ……………..(3)
Also, eqn. (2) passes through the point C (- 1, 3, 6).
∴ a (- 1 – 1) + b (3 – 2) + c (6 – 3) = 0
– 2a + b + 3c = 0 ……………………(4)
On solving eqn. (3) and eqn. (4) ; we have
\(\frac{a}{0+2}=\frac{b}{4-3}=\frac{c}{1-0}\)
i.e. \(\frac{a}{2}=\frac{b}{1}=\frac{c}{1}\) = k (say)
a = 2k ;
b = k ;
c = k
∴ eqn. (2) becomes ;
2k (x – 1) + k (y – 2) + k (z – 3) = 0
⇒ 2x + y + z – 7 = 0 ………………….(5)
Now any point on line (1) be
P (- t + 3, t – 4, 6t – 5)
Since the line (1) intersects the plane (5).
∴ P (- t + 3, t – 4, 6t – 5) lies on eqn. (5).
Thus, 2 (- t + 3) + t – 4 + 6t – 5 – 7 = 0
⇒ 5t – 10 = 0
⇒ t = 2
Thus, the required point of intersection of line (1) and plane (5) be
P(- 2 + 3, 2 – 4, 12 – 5)
i.e. P (1, – 2, 7).

Question 21.
A plane meets the coordinates axes at A, B and C respectively such that centroid of triangle ABC is (1, – 2, 3). Find the equation of the plane.
Solution:
Let the required eqn. of plane be
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\) = 1 ………………..(1)
Then the coordinates of A, B and C are
A (a, 0, 0), B (0, b, 0) and C (0, 0, c).
∴ centroid of the ∆ ABC are \(\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)\) also
given centroid of ABC are (1, – 2, 3)
i.e. \(\frac{a}{3}\) = 1
⇒ a = 3 ;
⇒ \(\frac{b}{3}\) = – 2
⇒ b = – 6
and \(\frac{c}{3}\) = 3
⇒ c = 9
putting the values of a, b, c in eqn. (1) ; we get
\(\frac{x}{3}+\frac{y}{-6}+\frac{z}{9}\) = 1
⇒ 6x – 3y + 2z = 18 be the required equation of plane.

Question 21 (old).
Find the coordinates of the point P where the line through A (3, – 4, – 5) and B (2, – 3, 1) crosses the plane passing through three points L (2, 2, 1), M (3, 0, 1) and N (4, – 1, 0). Also, find the rallo in which P divides the line segment AB.
Solution:
Eqn. of line through A (3, – 4, – 5) and B (2, – 3, 1) is given by
\(\frac{x-3}{2-3}=\frac{y+4}{-3+4}=\frac{z+5}{1+5}\)
i.e. \(\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+6}{6}\) ………………(1)
eqn. of any plane through point L (2, 2, 1) is given by
a (x – 2) + b (y – 2) + c (z – 1) = 0 ……………..(2)
Now the plane (2) passes through M (3, 0, 1) and N (4, – 1, 0).
i.e.a (3 – 2) + b (0 – 2) + c (1 – 1) = 0
⇒ a – 2b + 0c = 0 ………………..(3)
and a (4 – 2) + b (- 1 – 2) + c (0 – 1) = 0
⇒ 2a – 3b – c = 0 ………………(4)
On solving eqn. (3) and (4) ; we have
\(\frac{a}{2}=\frac{b}{1}=\frac{c}{1}\) = k (say)
∴ a = 2k ;
b = c = k
Thus from (2) ; we have,
2k (x – 2) + k (y – 2) + k (z – 1) = 0
⇒ 2x + y + z = 7 …………………..(5)
be the required eqn. of plane.
Any point on line (1) be
\(\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}\) = t (say)
The coordinates of any point P (- t + 3, t – 4, 6t – 5)
Clearly this point P lies in eqn. (5) ; we get
2 (- t + 3) + t – 4 + 6t – 5 = 7
⇒ 5t = 10
⇒ t = 2
∴ Coordinates of point P are (1, – 2, 7).
Let the point P divides the line segment AB in the ratio k : 1.

∴ The coordinates of point P are \(\left(\frac{2 k+3}{k+1}, \frac{-3 k-4}{k+1}, \frac{k-5}{k+1}\right)\)
Thus \(\frac{2 k+3}{k+1}\) = 1
⇒ 2k + 3 = k + 1
⇒ k = – 2
and \(\frac{-3 k-4}{k+1}\) = – 2
and \(\frac{k-5}{k+1}\) = 7
gives same value of k.
∴ required ratio be k : 1 i.e. – 2 : 1
i.e. 2 : 1 externally.

Question 22.
(i) Find the equation of the plane passing through the intersection of the planes x + 2y + 3z + 4 = 0 and x – y + z + 3 = 0 and passing through the origin.
(ii) Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0, x + y + z – 2 = 0 and the point (2, 2, 1).
Solution:
(i) The equation of any plane passing through the line of intersection of given planes
x + 2y + 3z + 4 = 0
and x – y + z + 3 = 0 is given by
(x + 2y + 3z + 4) + λ (x – y + z + 3) = 0
i.e. (1 + λ) x + (2 – λ) y + (3 + λ) z + 4 + 3λ = 0
Since the plane (1) passing through (0, 0, 0).
∴ (1 + λ) 0 + (2 – λ) 0 + (3 + λ) 0 + 4 + 3 = 0
λ = – \(\frac{4}{3}\)
putting the value of λ in eqn. (1); we have
(1 – \(\frac{4}{3}\)) x + (2 + \(\frac{4}{3}\)) y + (3 – \(\frac{4}{3}\)) z + – 4 = 0
⇒ – x + 10 y + 5z = 0
⇒ x – 10y – 5z = 0which is the required eqn. of plane.

(ii) The eqn. of any plane through the line of intersection of given planes
3x – y + 2z = 4 and x + y + z = 2 is given by
3x – y + 2z – 4 + λ (x + y + z – 2) = 0
⇒ (3 + λ) x + (- 1 + λ) y + (2 + λ) . 1 – 4 – 2λ = 0 …………………….(1)
Since plane (1) passes through (2, 2, 1).
∴ (3 + λ) 2 + (- 1 + λ) 2 + (2 + λ) . 1 – 4 – 2λ = 0
⇒ 3λ + 2 = 0
λ = – \(\frac{2}{3}\)
puing the value of λ = – \(\frac{4}{3}\) in eqn. (1) we have
7x – 5y + 4z = 8 be the required eqn. of plane.

Question 23.
(i) Find the equation of the plane passing through the intersection of the planes \(\vec{r} \cdot(\hat{i}+3 \hat{j}-\hat{k})\) and \(\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})\) = 3 and passing through the origin. (ISC 2018)
(iii) Find the equation of the plane passing through the intersection of the planes \(\vec{r} \cdot(2 \hat{i}+\hat{j}+3 \hat{k})\) = 7 and \(\vec{r} \cdot(2 \hat{i}+5 \hat{j}+3 \hat{k})\) = 9 and the point (2, 1, 3).
Solution:
(i) Given eqns. of planes be
\(\vec{r} \cdot(\hat{i}+3 \hat{j}-\hat{k})\) = 9 …………….(1)
and \(\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})\) = 3 …………….(2)
In cartesian form, eqns. of planes are given by
\((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+3 \hat{j}-\hat{k})\) = 9
⇒ x + 3y – z = 9 …………………(3)
and \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(2 \hat{i}-\hat{j}+\hat{k})\) = 3
⇒ 2x – y + z = 3 ……………………….(4)
Thus eqn. of plane passing through the intersection of planes (3) and (4) is given by
(x + 3y – z – 9) + k (2x – y + z – 3) = 0 …………………(5)
Since eqn. (5) passing through origin (0, 0, 0)
∴ (0 + 0 – 0 – 9) + k (0 – 0 + 0 – 3) = 0
⇒ – 9 – 3k = 0
⇒ k = – 3
putting the value of k in eqn. (5) ; we get
(x + 3y – z – 9) – 3 (2x – y + z – 3) = 0
⇒ – 5x + 6y – 4z = 0
⇒ 5x – 6y + 4z = 0, which is the required eqn. of plane.

(ii) The eqn. of plane passing throigh the lineof intersection of given planes be
\(\vec{r} \cdot(2 \hat{i}+\hat{j}+3 \hat{k})\) = 7
and \(\vec{r} \cdot(2 \hat{i}+5 \hat{j}+3 \hat{k})\) = 9
is given by \([\vec{r} \cdot(2 \hat{i}+\hat{j}+3 \hat{k})-7]+\lambda[\vec{r} \cdot(2 \hat{i}+5 \hat{j}+3 \hat{k})-9]\) = 0
\(\vec{r} \cdot[(2+2 \lambda) \hat{i}+(1+5 \lambda) \hat{j}+(3+3 \lambda) \hat{k}]\) – 7 – 9 λ = 0 …………………..(1)
Since the plane (1) passes through the point (2, 1, 2).
∴ \((2 \hat{i}+\hat{j}+3 \hat{k}) \cdot[(2+2 \lambda) \hat{i}+(1+5 \lambda) \hat{j}+(3+3 \lambda) \hat{k}]\) – 7 – 9 λ = 0
⇒ 2 (2 + 2λ) + (1 + 5λ) + 3 (3 + 3λ) – 7 – 9λ = 0
⇒ 9λ = – 7
⇒ λ = – \(\frac{7}{9}\)
putting the value of in eqn. (1); we get
\(\vec{r} \cdot\left[\frac{4}{9} \cdot \hat{i}-\frac{26}{9} \hat{j}+\frac{6}{9} \hat{k}\right]\) – 7 + 7 = 0
⇒ \(\vec{r} \cdot(2 \hat{i}-13 \hat{j}+3 \hat{k})\) = 0 be the required eqn. of plane.

Question 23 (old).
(ii) Find the vector equation of the plane passing through the interscction of the planes \(\vec{r} \cdot(2 \hat{i}-7 \hat{j}+4 \hat{k})\) and \(\vec{r} \cdot(3 \hat{i}-5 \hat{j}+4 \hat{k})\) + 11 = 0 and passing through the point (- 2, 1, 3).
Solution:
The vector eqn’s of given planes are \(\vec{r} \cdot(2 \hat{i}-7 \hat{j}+4 \hat{k})\) = 3
and \(\vec{r} \cdot(3 \hat{i}-5 \hat{j}+4 \hat{k})\) + 11 = 0
In cartesian form, eqn’s of planes are;
2x – 7y + 4z = 3 ………………..(1)
and 3x – 5y + 4z + 11 = 0 …………………(2)
Thus, equation of plane passing through the intersection of planes (1) and (2) is given by
(2x – 7y + 4z – 3) + λ (3x – 5y + 4z +11) = 0
Since eqn. (3) passing through the point (- 2, 1, 3).
∴ (- 4 – 7 + 12 – 3) + λ (- 6 – 5 +12 + 11) = 0
⇒ – 2 + λ (12) = 0
⇒ λ = \(\frac{1}{6}\)
putting the value of λ in eqn. (3) ; we have
(2x – 7y + 4z – 3) + \(\frac{1}{6}\) (3x – 5y + 4z + 11) = 0
⇒ 15x – 47y + 28z – 7 = 0 …………………(4)
If \(\vec{r}\) be the position vector of point P(x, y, z).
Then \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\)
Then eqn. (4) can be written as;
\((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(15 \hat{i}-47 \hat{j}+28 \hat{k})\) – 7 = 0
⇒ \(\vec{r} \cdot(15 \hat{i}-47 \hat{j}+28 \hat{k})\) – 7 = 0
which is the required vector eqn. of plane.

Question 24.
Find the equation of the plane passing through the line of intersection of the planes x + 2y + z = 3, 2x – y – z = 5 and the point (2, 1, 3). Also find direction ratios of a normal to this plane.
Solution:
Equations of given planes are
x + 2y + z – 3 = 0 ………………..(1)
and 2x – y – z – 5= 0 ………………..(2)
Thus, the equation of plane passing through the intersection of planes (1) and (2) is given by
(x + 2y + z – 3) + λ (2x – y – z – 5) = 0 ……………………(3)
where λ be the parameter.
Now eqn. (3) passes through the point (2, 1, 3).
∴ (2 + 2 + 3 – 3) + λ (4 – 1 – 3 – 5) = 0
⇒ 4 – 5λ = 0
⇒ λ = \(\frac{4}{5}\)
putting the value of in eqn. (3) ; we have
(x + 2y + z – 3) + \(\frac{4}{5}\) (2x – y – z – 5) = 0
⇒ 13x + 6y + z – 35 = 0, which is the required plane.
Direction ratios of the normal to plane are < 13, 6, 1 >.

Question 25.
Find the equation of the plane passing through the intersection of the planes 2x + 2y – 3z – 7 = 0 and 2x + 5y + 3z – 9 = 0 such that the intercepts made by the resulting plane on the x-axis and z-axis are equal. (ISC 2019)
Solution:
The equation of plane passing through
the point of intersection of given plane be given by
(2x + 2y – 3z – 7) + k (2x + 5y + 3z -9) = 0
⇒ (2 + 2k) x + (2 + 5k) y + (- 3 + 3k) z – 7 – 9k = 0
⇒ (2 + 2k) x + (2 + 5k) y + (- 3 + 3k) z = 7 + 9k …………………..(1)
For intercept made by plane (1) on x-axis ;
we put y = 0 = z in eqn. (1); we have
(2 + 2k) x₁ = 7 + 9k
⇒ x₁ = \(\frac{7+9 k}{2+2 k}\)
For intercept madeby plane (1) on z-axis
we put x = y = 0 in eqn. (1) ; we have
z₁ = \(\frac{7+9 k}{-3+3 k}\)
According to given condition,
x₁ = z₁
⇒ – 3 + 3k = 2 + 2k
[∵ k ≠ – \(\frac{7}{9}\)]
⇒ k = 2 + 3 = 5
putting the value of k = 5 in eqn. (1) ; we have
(2 + 10) x + (2 + 10) y + (- 3 + 15) z = 7 + 45
⇒ 12x + l2y + 12z = 52
which is the required equation of plane.