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

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

Question 1.
Find the ratio in which the join of the points P (2, 1, 5) and Q (3, 4, 3) is divided by the plane 2x + 2y – 2z – 1 = 0. Also find the coordinates of the point of division.
Solutions:
Let the point R divides the join of points P and Q in the ratio λ : 1.

∴ Coordinates of point R be \(\left(\frac{3 \lambda+2}{\lambda+1}, \frac{4 \lambda+1}{\lambda+1}, \frac{3 \lambda+5}{\lambda+1}\right)\)
Now point R lies on given plane i.e 2x + 2y – 2z – 1 = 0
\(2\left(\frac{3 \lambda+2}{\lambda+1}\right)+2\left(\frac{4 \lambda+1}{\lambda+1}\right)-2\left(\frac{3 \lambda+5}{\lambda+1}\right)\) – 1 = 0
⇒ 6λ + 4 + 8λ + 2 – 6λ – 10 – λ – 1 = 0
⇒ 7λ – 5 = 0
⇒ λ = \(\frac{5}{7}\)
Thus the required ratio be λ : 1 i.e. 5 : 7.
∴ Coordinates of point R are \(\left(\frac{\frac{15}{7}+2}{\frac{5}{7}+1}, \frac{\frac{20}{7}+1}{\frac{5}{7}+1}, \frac{\frac{15}{7}+5}{\frac{5}{7}+1}\right)\)
i.e. \(\left(\frac{29}{12}, \frac{27}{12}, \frac{50}{12}\right)\)
i.e. \(\left(\frac{29}{12}, \frac{9}{4}, \frac{25}{6}\right)\)

Question 2.
The cartesian equations of a line are 3x + 2 = 1 – 2y = z + 5. Find its direction numbers and also write dowia the vector equation of a line through (- 1, – 2, 0) and parallel to the given line.
Solution:
The cartesian equation of line be
3x + 2 = 1 – 2y = z + 5
⇒ \(\frac{x+\frac{2}{3}}{\frac{1}{3}}=-2\left(y-\frac{1}{2}\right)\) = z + 5
⇒ \(\frac{x+\frac{2}{3}}{\frac{1}{3}}=\frac{\left(y-\frac{1}{2}\right)}{-\frac{1}{2}}=\frac{z+5}{1}\)
i.e. \(\frac{x+\frac{2}{3}}{2}=\frac{y-\frac{1}{2}}{-3}=\frac{z+5}{1}\) ……………..(1)
∴ Direction numbers of line (1) are < 2, – 3, 1 >.
Thus vector equation of line passing through point with P.V
\(\vec{a}=-\hat{i}-2 \hat{j}+0 \hat{k}\) and parallel to vector
\(\vec{b}=2 \hat{i}-3 \hat{j}+\hat{k}\) is given by \(\)
\(\vec{r}=\vec{a}+\lambda \vec{b}\)
⇒ \(\vec{r}=-\hat{i}-2 \hat{j}+0 \hat{k}+\lambda(2 \hat{i}-3 \hat{j}+\hat{k})\)
where λ be the parameter.

Question 3.
Line L is drawn through the point C (1, 2, – 4) and parallel to the line through the points A (3, 2, 5) and B (3, 5, 9). Find two points on L which are at a distance of 10 units from C.
Solution:
Now Direction numbers of line through the points A (3, 2, 5) and B (3, 5, 9) are
< 3 – 3, 5 – 2, 9 – 5 > i.e.< 0, 3, 4 >
∴ eqn. of line through point C (1, 2, – 4) and to line having d’ ratios < 0, 3, 4 > is given by
\(\frac{x-1}{0}=\frac{y-2}{3}=\frac{z+4}{4}\) = t (say)
So Any point on line (1) be P (1, 3t + 2, 4t – 4)
It is given that |CP| = 10 units
\(\sqrt{(1-1)^2+(3 t+2-2)^2+(4 t-4+4)^2}\) = 10
⇒ \(\sqrt{9 t^2+16 t^2}\) = 10 ;
On squaring
⇒ 25t² = 10
⇒ t² = 4
⇒ t = ± 2
When t = 2, any point on line L be P (1, 8, 4)
When t = – 2, any point on line L be Q (1, – 4, – 12)

Question 4.
If a line makes angles α, β, γ, δ with the four diagonals of a cube, prove that sin² α + sin² β + sin² γ + sin² δ = \(\frac{8}{3}\).
Solution:
Taking O as origin.
The coterminous edges of cube i.e. OA, OB and OC as three coordinate axes.
Let the length of each side of cube be a units.
Thus the coordinates of vertices of cube are O (0, 0, 0);
A (a, 0, 0), B (0, a, 0); C (0, 0, a); p (a, a, a) ; L (a, 0, a) ; M (a, a, 0) and N (0, a, a).
The four diagonals of cube are OP, AN, BL and CM.

Thus direction ratios of OP are < a – 0, a – 0,a – 0 > i.e. < 1, 1, 1 >
and The direction ratios of AN, BL and CM are
< 0 – a, a – 0, a – 0 > ;
i.e. < – 1, 1, 1 > ; < 1, – 1, 1 > and < 1, 1,- 1 >
∴ D’ cosines of four diagonals OP, AN, BL and CM are
< \(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}>;<-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}>;<\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\) >
and < \(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\) >
Let < l, m, n > be the direction cosines of given line.
Let α, β, γ and δ are the angles made by line with diagonals OP, AN, BL and CM respectively.
∴ cos α = \(l \cdot \frac{1}{\sqrt{3}}+m \cdot \frac{1}{\sqrt{3}}+n \cdot \frac{1}{\sqrt{3}}\)
= \(\frac{(l+m+n)}{\sqrt{3}}\) …………………(1)
cos β = \(\frac{-l+m+n}{\sqrt{3}}\) ……………………..(2)
cos γ = \(\frac{l-m+n}{\sqrt{3}}\) ……………………..(3)
and cos δ = \(\frac{l+m-n}{\sqrt{3}}\) ……………………..(4)
∴ cos² α + cos² β + cos² γ + cos² δ = \(\frac{(l+m+n)^2}{3}+\frac{(-l+m+n)^2}{3}+\frac{(l-m+n)^2}{3}+\frac{(l+m-n)^2}{3}\)
= \(\frac{1}{3}\) [4l² + 4m² + 4n²]
[∵ l² + m² + n² = 1]
= \(\frac{4}{3}\) . 1
⇒ (1 – sin² α) (1 – sin² β) + (1 – sin² γ) + (1 – sin² δ) = \(\frac{4}{3}\)
⇒ sin² α + sin² β + sin² γ + sin² δ = 4 – \(\frac{4}{3}\)
= \(\frac{8}{3}\)

Question 5.
A (1, 0, 4), B (0, – 11, 3), C (2, – 3, 1) are three points and D is the foot of perpendicular from A on BC. Find the coordinates of D.
Solution:

Let D be the foot of ⊥ drawn from A (1, 0, 4) on line BC and D divides BC in the ratio λ : 1.
∴ Coordinates of D are \(\left(\frac{2 \lambda+0}{\lambda+1}, \frac{-3 \lambda-11}{\lambda+1}, \frac{\lambda+3}{\lambda+1}\right)\)
∴ d’ ratios of line AD are \(\left(\frac{2 \lambda}{\lambda+1}-1, \frac{-3 \lambda-11}{\lambda+1}, \frac{\lambda+3}{\lambda+1}-4\right)\)
i.e. \(\left(\frac{\lambda-1}{\lambda+1}, \frac{-3 \lambda-11}{\lambda+1}, \frac{-3 \lambda-1}{\lambda+1}\right)\)
andd’ ratiosoflineBCare < 2 – 0, – 3 + 11, 1 – 3 > i.e. < 2, 8, – 2 > i.e. < 1, 4, – 1 >
Since line AD is ⊥ to line BC.
∴ \(\left(\frac{\lambda-1}{\lambda+1}\right) 1+\left(\frac{-3 \lambda-11}{\lambda+1}\right) 4+\left(\frac{-3 \lambda-1}{\lambda+1}\right)(-1)\) = 0
⇒ λ – 1 – 12λ – 44 + 3λ + 1 = 0
⇒ – 8λ – 44 = 0
⇒ λ = – \(\frac{44}{8}\)
= – \(\frac{11}{2}\)
∴ Coordinates of D are \(\left(\frac{-11}{-\frac{11}{2}+1}, \frac{\frac{33}{2}-11}{-\frac{11}{2}+1}, \frac{-\frac{11}{2}+3}{-\frac{11}{2}+1}\right)\)
i.e. \(\left(\frac{22}{9},-\frac{11}{9}, \frac{5}{9}\right)\)

Question 6.
Find the image of the point p (5, 9, 3) in the line \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\).
Solution:
Given eqn. of line be \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) ……………….(1)
Let N be the foot of ⊥ drawn from P (5, 9, 3) on given line (1).
So any point on line (1) be (2t + 1, 3t + 2, 4t + 3)
we want to find the value of t for which this point becomes N.

∴ D’ ratios of line PN are < 2t + 1 – 5, 3t + 2 – 9, 4t + 3 – 3 > i.e. < 2t – 4, 3t – 7, 4t >
Also D’ ratios of given line (1) are < 2, 3, 4 >
Since line PN be ⊥ to line (1).
∴ (2t – 4) 2 + (3t – 7) 3 + 4t × 4 = 0
⇒ 4t – 8 + 9t – 21 + 16t
⇒ t = 1
∴ Coordinates of point N be (3, 5, 7).
Let Q (α, β, γ) be the image of point P (5, 9, 3) then N be the mid point of PQ.
∴ \(\frac{\alpha+5}{2}\) = 3 ;
\(\frac{\beta+9}{2}\) = 1 ;
\(\frac{\gamma+3}{2}\) = 7
⇒ α = 6 – 5
α = 1
β = 10 – 9 = 1 ;
γ = 14 – 3 = 11
Thus the required image of given point be (1, 1, 11).

Question 7.
Find whether or not the two lines given below intersect :
\(\vec{r}=(2 \lambda+1) \hat{i}-(\lambda+1) \hat{j}+(\lambda+1) \hat{k}\),
\(\vec{r}=(3 \mu+2) \hat{i}-(5 \mu+5) \hat{j}+(2 \mu+1) \hat{k}\).
Solution:
Given lines can be written as
\(\vec{r}=(2 \lambda+1) \hat{i}-(\lambda+1) \hat{j}+(\lambda+1) \hat{k}\),
and \(\vec{r}=(3 \mu+2) \hat{i}-(5 \mu+5) \hat{j}+(2 \mu+1) \hat{k}\)
Its cartesian eqn’s are;
\(\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-1}{1}\) = t (say) ………………..(3)
and \(\frac{x-2}{3}=\frac{y+5}{-5}=\frac{z-1}{2}\) = s (say) ……………….(4)
So any point on line (3) be M (2t + 1, – 1 – 1, t + 1)
it will lies on eqn. (4)
if \(\frac{2 t+1-2}{3}=\frac{-t-1+5}{-5}=\frac{t+1-1}{2}\) are consistent
¡if (2t – 1) (- 5) = 3 (- t + 4)
and 2 (- t + 4) = – 5t are consistent
iff – 7t = 7
and 3t = – 8 are consistent
iff t = – 1
and t = – \(\frac{8}{3}\) are consistent
Hence given lines do not intersects.

Aliter:
On comparing given lines with

Thus shortest distance between given lines is non-zero.
Hence lines do not intersects with each other.

Question 8.
Define the line of shortest distance between the two lines. Find the shortest distance and the vector equation of the line of shortest distance between the lines given by :
\(\vec{r}=(-4 \hat{i}+4 \hat{j}+\hat{k})+\lambda(\hat{i}+\hat{j}-\hat{k})\) and \(\vec{r}=(-3 \hat{i}-8 \hat{j}-3 \hat{k})+\mu(2 \hat{i}+3 \hat{j}+3 \hat{k})\).
Solution:
Line of shortest distance:
If l₁ and l₂ are two skew lines and then their is one and only one line which is ⊥ to both l₁ and l₂ and this line is called line of shortest distance.
The equation of given lines are ;
\(\vec{r}=(-4 \hat{i}+4 \hat{j}+\hat{k})+\lambda(\hat{i}+\hat{j}-\hat{k})\)
and \(\vec{r}=(-3 \hat{i}-8 \hat{j}-3 \hat{k})+\mu(2 \hat{i}+3 \hat{j}+3 \hat{k})\)
The cartesian equations of given lines are ;
\(\) ……………….(1)
and \(\) ……………………(2)
So any point on line (1) be M (t – 4, t + 4, – t + 1)
and any point on line (2) be N (2s – 3, 3s – 8, 3s – 3)
∴ Direction ratios of line MN are < 2s – 3 – t + 4, 3s – 8 – t – 4, 3s – 3 + t – 1 >
i.e. < 2s – t + 1, 3s- t – 12, 3s + t – 4 >
So MN be the shortest distance between lines (1) and (2) and line ⊥ to (1) and (2).
∴ (2s – t + 1) 1 + (3s – t – 12) 1 + (3s + t – 4) (- 1) = 0
⇒ 2s – 3t = 7 ………………………(3)
and (2s – t + 1) 2 + (3s – t – 12) 3 + (3s + t – 4) 3 = 0
⇒ 22s – 21 = 46 …………………(4)
On solving (3) and (4) ; we have
t = – 1 ;s = 2
Thus t = – 1 gives M (- 5, 3, 2)
and s = 2 gives N (1, – 2, 3)
∴ S.D between given lines = |MN|
= \(\sqrt{(1+5)^2+(-2-3)^2+(3-2)^2}\)
= \(\sqrt{36+25+1}=\sqrt{62}\) units
and eqn. of line of S.D i.e. MN is given by
\(\frac{x+5}{1+5}=\frac{y-3}{-2-3}=\frac{z-2}{3-2}\)
i.r. \(\frac{x+5}{6}=\frac{y-3}{-5}=\frac{z-2}{1}\)
Hence, vector eqn. of line of S.D is given by
\(\vec{r}=-5 \hat{i}+3 \hat{j}+2 \hat{k}+\lambda(6 \hat{i}-5 \hat{j}+\hat{k})\)
where λ be the parameter.

Question 9.
Find the vector equation of the plane which ¡s normal to the vector \(\hat{i}+\hat{j}+\hat{k}\) and passes through the point (2, 3, 4). Also, find its cartesian equation.
Solution:
Given \(\vec{n}=\hat{i}+\hat{j}+\hat{k}\)
and \(\vec{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}\)
So vector equation of plane which is normal to vector \(\hat{i}+\hat{j}+\hat{k}\) and passes through the point (2, 3,4) is given by
\((\vec{r}-\vec{a}) \cdot \vec{n}\) = 0
⇒ \(\vec{r} \cdot \vec{n}=\vec{a} \cdot \vec{n}\)
\(\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k})\)
= 2(1) + 3(1) + 4(1) = 9 …………………(1)
If \(\vec{r}\) be the position vector of point P (x, y z)
Then \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\)
∴ from (1) ; we have
\((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k})\) = 9
⇒ x + y + z = 9
which is the required cartesian eqn. of plane.

Question 9 (old).
Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.
Solution:
eqn. of ZOX plane be y= 0
∴ eqn. of plane parallel to ZOX plane be
y = k …………………..(1)
Since ¡lis given that plane cut off intercept 3 on y-axis.
∴ plane (1) passes through (0, 3, 0).
∴ 3 = k
Thus required eqn. of plane be y = 3.

Question 10.
The line drawn from (4, – 1, 2) to the point (- 3, 2, 3) meets a plane at right angles at the point (- 10, 5, 4). Find the vector equation of the plane.
Solution:
Direction ratios of the line normal to plane are
< – 3 – 4, 2 + 1, 3 – 2 > i.e. < – 7, 3, 1 >.
∴ \(\vec{n}\) = vector along normal to plane
= \(-7 \hat{i}+3 \hat{j}+\hat{k}\)
and \(\vec{a}=-10 \hat{i}+5 \hat{j}+4 \hat{k}\)
Thus, required vector eqn. of plane be,
\(\vec{r} \cdot \vec{n}=\vec{a} \cdot \vec{n}\)
∴ \(\vec{r} \cdot(-7 \hat{i}+3 \hat{j}+\hat{k})\) = \((-10 \hat{i}+5 \hat{j}+4 \hat{k})(-7 \hat{i}+3 \hat{j}+\hat{k})\)
⇒ \(\vec{r} \cdot(-7 \hat{i}+3 \hat{j}+\hat{k})\) = – 10 (- 7) + 5 (3) + 4 (1)
= 70 + 15 + 4 = 89
which is the required vector eqn. of plane.

Question 11.
Find the equation of the plane passingthrough the point (1, 2, 1) and perpendicular to the line joining the points (1, 4, 2) and (2, 3, 5). FInd also the perpendicular distance of the origin from this plane.
Solution:
D’ ratios of the line joining (1, 4, 2) and (2, 3, 5) are
< 2 – 1, 3 – 4, 5 – 2 > i.e.< 1, – 1, 3 >
∴ eqn. of plane through (1, 2, 1) and ⊥ to line joining (1, 4, 2) and (2, 3, 5) is given by
1 (x – 1) – 1 (y – 2) + 3 (z – 1) = 0
⇒ x – y + 3z – 2 = 0
⇒ x – y + 3z = 2 …………………(1)
⇒ \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}-\hat{j}+3 \hat{k})\) = 2
⇒ \(\vec{r} \cdot(\hat{i}-\hat{j}+3 \hat{k})\) = 2 is the required vector eqn. of plane.
∴ Required ⊥distance of the origin from plane (1)
= \(\frac{|0-0+3(0)-2|}{\sqrt{1^2+(-1)^2+3^2}}\)
= \(\frac{2}{\sqrt{11}}\) units

Question 12.
Find the vector equation of the line passing through the point (3, 1, 2) and perpendicular to the plane \(\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})\) = 4. Also find the point of intersection of this line and the plane.
Solution:
Given eqn. of plne be \(\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})\) = 4
In cartesian eqn. of plane be
2x – y + z – 4 = 0 ……………….(1)
∴ D’ numbers of normal to plane (1) are < 2, – 1, 1 >
∴ eqn. of line passing through the point (3, 1, 2)
and ⊥ to plane (1) i.e. having direction ratios < 2, – 1, 1 > is given by
\(\frac{x-3}{2}=\frac{y-1}{-1}=\frac{z-2}{1}\) = t (say) ………………….(2)
Any point on line (2) be (2t + 3, t + 1, t + 2)
For the point of intersection of line (2) and plane (1), this point must lies on plane (I).
∴ 2 (2t + 3) – (- t + 1) + t + 2 – 4 = 0
⇒ 6t + 3 = 0
⇒ t = – \(\frac{1}{2}\)
∴ required point of intersection of line (1) and plane (2) be
\(\left(-1+3, \frac{1}{2}+1,-\frac{1}{2}+2\right)\)
i.e. \(\left(2, \frac{3}{2}, \frac{3}{2}\right)\)
Thus vector eqn. of line (2) is given by
where λ be the parameter.

Question 13.
Find the vector equation of the plane that bisects the line joining the points A and B with position vectors \(-\hat{i}+2 \hat{j}+3 \hat{k}\) and \(3 \hat{i}-5 \hat{j}+6 \hat{k}\) at right angles .
Solution:
∴ D’ ratios of line joining A (- 1, 2, 3) and B (3, – 5, 6) are
< 3 – (- 1), – 5 – 2, 6 – 3 >
i.e. < 4, – 7, 3 >

also mid-point of AB is given by
M \(\left(\frac{-1+3}{2}, \frac{2-5}{2}, \frac{3+6}{2}\right)\)
i.e. M \(\left(1, \frac{-3}{2}, \frac{9}{2}\right)\)
∴ eqn. of plane through \(\left(1,-\frac{3}{2}, \frac{9}{2}\right)\) is given by
a (x – 1) + b (y + \(\frac{3}{2}\)) + c (z – \(\frac{9}{2}\)) = 0 …………….(1)
where < a, b, c > are the direction ratios of normal to plane.
Since the line is ⊥ to the plane (1)
∴ Normal to plane is || to line AB
i.e. \(\frac{a}{4}=\frac{b}{-7}=\frac{c}{3}\) = k (say) ; where k ≠ 0
i.e. a = 4k ;
b = – 7k ;
c = 3k
∴ eqn. (1) becomes ;
4 (x – 1) – 7 (y + \(\frac{3}{2}\)) + 3 (z – \(\frac{9}{2}\)) = 0
⇒ 4x – 7y + 3z – 4 – \(\frac{21}{2}-\frac{27}{2}\) = 0
⇒ 4x – 7y + 3z = 28 is the required eqn. of plane.
Thus required vector eqn. of plane be
\(\vec{r} \cdot(4 \hat{i}-7 \hat{j}+3 \hat{k})\) = 28
where \(\vec{r}\) be the P.V. of any point P (x, y, z) on plane.

Question 14.
Find the vector equation of the plane passing through the point \(3 \hat{i}+4 \hat{j}+2 \hat{k}\) and parallel to the vectors \(\hat{i}+2 \hat{j}+3 \hat{k}\) and \(\hat{i}-\hat{j}+\hat{k}\).
Solution:
The eqn. of plane through the point \(3 \hat{i}+4 \hat{j}+2 \hat{k}\) is given by
a (x – 3) + b (y – 4) + c (z – 2) = 0 ……………….(1)
where < a, b, c > are the direction numbers of normal to plane (1).
Since the plane (1) is parallel to vector \(\hat{i}+2 \hat{j}+3 \hat{k}\).
∴ normal to plane (1) is ⊥ to vector \(\hat{i}+2 \hat{j}+3 \hat{k}\).
∴ a + 2b + 3c = 0 ……………..(2)
Since the plane (1) is parallel to vector \(\hat{i}-\hat{j}+\hat{k}\).
∴ normal to plane (1) is ⊥ to vector \(\hat{i}-\hat{j}+\hat{k}\)
a – b + c = 0 …………………(3)
On solving eqn. (2) and eqn. (3) ; we have
\(\frac{a}{2+3}=\frac{b}{3-1}=\frac{c}{-1-2}\)
i.e. \(\frac{a}{5}=\frac{b}{2}=\frac{c}{-3}\) = k (say) ; where k ≠ 0
a = 5k ;
b = 2k
and c = – 3k
∴ eqn. (1) becomes ;
5k (x – 3) + 2k (y – 4) – 3k (z – 2) = 0
⇒ 5x + 2y – 3z – 17 = 0 [∵ k ≠ 0]
If be the position vector of P x, y, z) on plane (4).
∴ eqn. (4) becomes ;
\(\vec{r} \cdot(5 \hat{i}+2 \hat{j}-3 \hat{k})\) = 17
which is the required vector equation of plane.

Question 15.
A variable plane passes through a fixed point (α, β, γ) and meets the coordinate axes in A, B and C respectively. Find the locus of the point common to the planes through A, B, C and parallel to the coordinate planes.
Solution:
Let the equation of plane be
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\) = 1 ………………(1)
Since plane (1) meets the coordinate axes at A, B and C.
For intersection of plane (1) with x-axis, we have
y = z = 0
∴ from (1) ;
\(\frac{x}{a}\) = 1
⇒ x = a
∴ point of intersection of plane (I) with x-axis at A (a, 0, 0)
Similarly point of intersection of plane (1)
with y-axis and z-axis at B (0, b, 0) and C (0, 0, c).
Since plane (1) passes through (α, β, γ)
∴ \(\frac{\alpha}{a}+\frac{\beta}{b}+\frac{\gamma}{c}\) = 1 ……………….(2)

Now eqn.of plane || to XOY plane be z = k
and it passes through C (0, 0, c)
∴ k = c
∴ z = c
Similarly eqn. of plane || to YOZ and ZOX
plane passes through A (a, 0, 0) and B(0, b, 0) are given by
x = a
and y = b.
∴ from (2) ; we have
\(\frac{\alpha}{x}+\frac{\beta}{y}+\frac{\gamma}{z}\) = 1,
which gives the required locus of the point common to the planes through A, B, C and parallel to the coordinate planes.

Question 16.
If L, M, N are the feet of perpendiculars drawn from the point P (3, 7, 9) on the coordinate planes, find the equation of the planes passing through the points L, M, N.
Solution:
eqn. of XOY plane be z = 0 ……………….(1)
D’ numbers of normal to plane(1) are < 0, 0, 1 >
∴ eqn. of line PL is given by
\(\frac{x-3}{0}=\frac{y-7}{0}=\frac{z-9}{1}\) ……………….(2)

Any point on line (2) is given by L (3, 7, t + 9)
and this point L lies on plane (1).
∴ t + 9 = 0
⇒ t = – 9
∴ point L becomes (3, 7, 0)
eqn. of YOZ plane be x = 0 ………………..(3)
and direciion ratios of normal to plane (3) are < 1, 0, 0 >.
∴ eqn. of line PM is given by
\(\frac{x-3}{1}=\frac{y-7}{0}=\frac{z-9}{0}\) …………………….(4)

Any point on line (4) is given by M (s + 3, 7, 9)
and this point lies on plane (3)
∴ s + 3 = 0
⇒ s = – 3
∴ point M becomes (0, 7, 9).
Similarly eqn. of ZOX plane be y = 0 …………………(5)
and direction ratios of normal to plane (2) are < 0, 1, 0 >
∴ eqn. of line through P and ⊥ to plane (5)
i.e. PN is given by
\(\frac{x-3}{0}=\frac{y-7}{1}=\frac{z-9}{0}\) ………………….(6)

So any point on line (6) be N (3, r + 7, 9)
and this point lies on eqn. (5)
∴ r + 7 = 0
⇒ r = – 7
∴ point N becomes (3, 0, 9).
We know that, eqn. of plane through (x₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃, z₃) is given by
\(\left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
x_3-x_1 & y_3-y_1 & z_3-z_1
\end{array}\right|\) = 0
∴ eqn. of plane passing through the points L (3, 7, 0) ; M (0, 7, 9) and N (3, 0, 9) is given by
\(\left|\begin{array}{ccc}
x-3 & y-7 & z-0 \\
0-3 & 7-7 & 9-0 \\
3-3 & 0-7 & 9-0
\end{array}\right|\) = 0
i.e. \(\left|\begin{array}{ccc}
x-3 & y-7 & z \\
-3 & 0 & 9 \\
0 & -7 & 9
\end{array}\right|\) = 0
Expanding along R₁ ; we have
⇒ (x – 3) (0 + 63) – (y – 7) (- 27) + z (21) = 0
⇒ 21 (x – 3) + 9 (y – 7) + 7z = 0
⇒ 21x + 9y + 7z – 126 = 0
which is the required eqn. of plane.

Question 17.
Find the equation of the plane through the point (4, – 3, 2) and perpendicular to the line of intersection of the planes x -y + 2z – 3 = 0 and 2x – y – 3z = 0.
Solution:
The eqn. of plane through the point (4, – 3, 2) is given by
a (x – 4) + b (y + 3) + c (z – 2) = 0 …………………………(1)
where < a, b, c > are the direction numbers of normal to plane (1).
Since plane (1) is ⊥ to both planes i.e.
x – y + 2z – 3 = 0
and 2x – y – 3z = 0
∴ a – b + 2c = 0 ……………….(2)
and 2a – b – 3c = 0
On solving eqn. (2) and eqn. (3) ; we have
\(\frac{a}{3+2}=\frac{b}{4+3}=\frac{c}{-1+2}\)
i.e. \(\frac{a}{5}=\frac{b}{7}=\frac{c}{1}\) = k (say) where k ≠ 0
∴ a = 5k ; b = 7k ; c = k
putting these values of a, b, c in eqn. (1) ; we have
5k (x – 4) + 7k (y + 3) + k (z – 2) = 0
⇒ 5x + 7y + z – 1 = 0
which is the required cartesian equation of plane.

Question 18.
Find the coordinates of the foot of perpendicular from the point P (2, 3, 7) to the plane 3x – y – z = 7. Also find the length of perpendicular and the image of the Point P in the given plane.
Solution:
Let the given point be p (2, 3, 7)
and given plane be 3x – y – z – 7 = 0 ……………….(1)
∴ eqn. of line PM is
\(\frac{x-2}{3}=\frac{y-3}{-1}=\frac{z-7}{-1}\) ………………..(2)

Any point on line (2) is (3t + 2, – t + 3, – t + 7)
Now this point is M if it ¡les on (1)
i.e. 3 (3t + 2) – (- t + 3) – (- t + 7) – 7 = 0
⇒ 11t – 11 = 0
⇒ t = 1
∴ Required coordinates of foot of L is M (5, 2, 6).
∴ length of. ⊥ from P (2, 3, 7) to plane (1) = |PM|
= \(\sqrt{(5-2)^2+(2-3)^2+(6-7)^2}\)
= \(\sqrt{9+1+1}=\sqrt{11}\)
Let Q (α, β, γ) be the image of point P if M be the mid-point of PQ.
∴ \(\frac{\alpha+2}{2}\) = 5 ;
\(\frac{\beta+3}{2}\) = 2 ;
and \(\frac{\gamma+7}{2}\) = 6
i.e. α = 8 ;
β = 1
and γ = 5
Thus the required image of point Pin given plane be Q (8, 1, 5).

Question 19.
Find the equation of the plane which makes intercepts – 3 and 5 on x-axis and z-axis respectively and is parallel to y-axis.
Solution:
Since the plane makes intercepts – 3 and 4 on x-axis and z-axis respectively.
∴ required plane passes through (- 3, 0, 0) and (0, 0, 4).
Let the equation of plane through the point (- 3, 0, 0) be
a (x + 3) + by + cz = 0 ………………(1)
Since eqn. (1) i.e. plane (1) passes through (0, 0, 4)
∴ a (0 + 3) + b × 0 + 4c = 0
i.e. 3a + 0b + 4c = 0 ………………(2)
Also plane (1) is parallel toy-axis
∴ normal to plane is .1 toy-axis whose cf ratios are < 0, 1, 0 >
∴ 0a + 1b + 0c = 0 ………………..(3)
On solving eqn. (2) and eqn. (3); we have
\(\frac{a}{-4}=\frac{b}{0}=\frac{c}{3}\) = k (say) ; k ≠ 0
∴ a = – 4k ;
b = 0 ;
c = 3k
putting the values of a, b and c in eqn. (1);
– 4k (x + 3) + 0 × y + 3kz = 0
⇒ – 4x + 3z – 12 = 0
⇒ 4x – 3z + 12 = 0
which is the required eqn. of plane.

Question 21.
Find the equation of the plane passing through the line 3x – 2y + z + 5 = 0 = 2x + 3y + 4z – 4 and parallel to the line \(\frac{x-2}{3}=\frac{y+1}{5}=\frac{z+11}{-2}\).
Solution:
The equation of plane through the line 3x-2y + z + 5 = 0 = 2x + 3y + 4z – 4 is given by
(3x – 2y + z + 5) + λ (2x + 3y+ 4z – 4) = 0
⇒ (3 + 2λ) x + (- 2 + 3λ) y + (1 + 4λ) + 5 – 4λ = 0 ……………….(1)
∴ Direction ratios of norma! to plane (1) are
< 3 + 2λ, – 2 + 3λ, 1 + 4λ >
Also eqn. of given line be
\(\frac{x-2}{3}=\frac{y+1}{5}=\frac{z+11}{-2}\) ………………..(2)
Since the plane (1) is parallel to line (2)
∴ normal to plane (1) is ⊥ to line (2).
⇒ (3 + 2λ) 3 + (- 2 + 3λ) 5 + (1 + 4λ) (- 2) = 0
⇒ 9 + 6λ – 10 + 15λ – 2 – 8λ = 0
⇒ 13λ – 13 = 0
⇒ λ = \(\frac{13}{13}\)
putting λ = \(\frac{13}{13}\) in eq. (1) ; we have
\(\left(3+\frac{6}{13}\right) x+\left(-2+\frac{9}{13}\right) y+\left(1+\frac{12}{13}\right) z+5-\frac{12}{13}\) = 0
⇒ 45x – 17y + 25z + 53 = 0, which is the required eqn. of plane.

Question 21.
Find the equation fo 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:
The eqns. of given planes are
x + 2y + z = 3 …………………(1)
and 2x – y – z = 5
∴ The required equation of plane through the line of intersection of planes (I) and (2) is given by
(x + 2y + z – 3) + λ (2x – y – z – 5) = 0 …………………(3)
It is given that plane (3) is passes through the point (2, 1, 3).
∴ (2 + 2 + 3 – 3) + (4 – 1 – 3 – 5) = 0
⇒ 4 – 5λ = 0
⇒ λ = \(\frac{4}{5}\)
putting the value of in qn. (3) ; we have
(x + 2y + z – 3) + \(\frac{4}{5}\) (2x – y – z – 5) = 0
⇒ 13x + 6y + z – 35 = 0, which is the required eqn. of plane.
Also, the direction ratios of the normal to plane are < 13, 6, 1 >.

Question 22.
Find the equation of the plane passing through the line of intersection of the planes x + 3y – 2z + 1 = 0 and 2x – 3y + 5z + 3 = 0 and which cuts off intercepts equal ¡n magnitude but opposite in sign on x-axis and z-axis.
Solution:
The required eqn. of plane passing through the line of intersection of both given planes
x + 3y – 2z + 1 = 0
and 2x – 3y + 5z + 3 = 0 is given by
(x + 3y – 2z + 1) + (2x – 3y + 5z + 3) = 0 ……………………….(1)
For intercept of plane (1) with x-axis, we have y = 0 = z
∴ from (1) ;
(x + 1) + λ (2x + 3)O = 0
⇒ (1 + 2λ) x + 1 + 3λ = 0
⇒ x = \(-\frac{(1+3 \lambda)}{1+2 \lambda}\)
For intercept of plane (1) with z-axis, we have x = 0 = y
∴ from (1) ;
(- 2z + 1) + λ (5z + 3) = 0
⇒ (- 2 + 5λ) z + 1 + 3λ = 0
⇒ z = \(\frac{1+3 \lambda}{2-5 \lambda}\)
according to given condition, we have
\(-\frac{(1+3 \lambda)}{1+2 \lambda}=-\frac{1+3 \lambda}{2-5 \lambda}\)
⇒ (1 + 3λ) (2 – 5λ – 1 – 2λ) = 0
⇒ (1 + 3λ) (1 – 7λ) = 0
⇒ λ = – \(\frac{1}{3}\), \(\frac{1}{7}\)
But at λ = – \(\frac{1}{3}\), both x-intercept and z-intercept is 0.
∴ given condition is not satisfied
∴ λ = + \(\frac{1}{7}\)
putting λ = \(\frac{1}{7}\) in eqn.(1) ; we have
(x + 3y – 2z + 1) + \(\frac{1}{7}\) (2x – 3y + 5z + 3) = 0
⇒ 9x + 18y – 9z + 10 = 0, which isthe required eqn. of plane.

Question 23.
Find the vector equation of the plane pssing through the points \(-2 \hat{i}+6 \hat{j}-6 \hat{k}\), \(-3 \hat{i}+7 \hat{j}-9 \hat{k}\) and \(-5 \hat{i}-6 \hat{k}\). Also find the length of perpendicular from origin to this plane.
Solution:
Here, we want to find the eqn. of plane through the points (- 2, 6, – 6) ; (- 3, 7, – 9) and (- 5, 0, – 6).
We know that, eqn. of plane through the points (x₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃, z₃) is given by
\(\left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
x_3-x_1 & y_3-y_1 & z_3-z_1
\end{array}\right|\) = 0

i.e \(\left|\begin{array}{ccc}
x+2 & y-6 & z+6 \\
-3+2 & 7-6 & -9+6 \\
-5+2 & 0-6 & -6+6
\end{array}\right|\) = 0

i.e. \(\left|\begin{array}{ccc}
x+2 & y-6 & z+6 \\
-1 & 1 & -3 \\
-3 & -6 & 0
\end{array}\right|\) = 0
Expanding along R₁ ; we have
(x + 2 × 0 – 18) – (y – 6 × 0 – 9) + (z + 6) (6 + 3)=0
⇒ – 18 (x + 2) + 9 (y – 6) + 9 (z + 6) = 0
⇒ – 2 (x + 2) + y – 6 + z + 6 = 0
⇒ – 2x + y + z – 4 = 0
⇒ 2x – y – z + 4 = 0
If \(\vec{r}\) be the P.V. of point (x, y, z) on plane (1)
∴ \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\)
Thus eqn. (1) becomes ;
\((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(2 \hat{i}-\hat{j}-\hat{k})\) + 4 = 0
⇒ \(\vec{r} \cdot(2 \hat{i}-\hat{j}-\hat{k})\) + 4 = 0
which is the vector eqn. of plane.
∴ length of ⊥ from (0, 0, 0) on plane (1) = \(\frac{|2(0)-0-0+4|}{\sqrt{2^2+(-1)^2+(-1)^2}}\)
= \(\frac{4}{\sqrt{6}}\) units

Question 25.
Find the coordinates of the point where the line through (3, – 4, – 5) and (2, – 3, 1) crosses the plane passing through three points (2, 2, 1), (3, 0, 1) and (4, – 1, 0). (NCERT Exemplar)
Solution:
The equation 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)
Now eqn. of plane through the point (2, 2, 1) is given by
a (x – 2) + b (y – 2) + c (z – 1) = 0 ……………….(2)
where < a, b, c > are the direction ratios of normal to plane (2).
Since plane (3) passes through (3, 0, 1).
∴ a (3 – 2) + b(0 – 2) + c (1 – 1) = 0
i.e. a – 2b + 0c = 0 …………………..(3)
Also plane (2) passes through (4, – 1, 0).
∴ a (4 – 2) + b (- 1 – 2) + c (0 – 1) = 0
i.e. 2a – 3b – c = 0 …………………..(4)
On solving eqn. (3) and eqn. (4); we have
\(\frac{a}{2-0}=\frac{b}{0+1}=\frac{c}{-3+4}\)
i.e. \(\frac{a}{2}=\frac{b}{1}=\frac{c}{1}\) = k (say); where k ≠ 0
∴ a = 2k ;
b = k ;
c = k
putting the values of a, b, c and in eqn. (2); we get
2k (x – 2) + k (y – 2) + k (z – 1) = 0
⇒ 2x + y + z – 7 = 0 ………………..(5)
[∵ k ≠ 0]
which gives the rquired eqn. of plane.
So any point on line (1) be P (- t + 3, t – 4, 6t – 5).
For point of intersection of line (1) and plane (5), the point P lies on plane (5).
∴ 2(- t + 3) + t – 4 + 6t – 5 – 7 = 0
⇒ 5t – 10 = 0
⇒ t = 2
∴ required point of intersection of line (1) and plane (5) be given by (- 2 + 3, 2 – 4, 12 – 5)
i.e. (1, – 2, 7).

Question 25.
Find the image of the point \(2 \hat{i}+3 \hat{j}-4 \hat{k}\) in the plane \(\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})\) = 3.
Solution:
Cartesian eqn. of given plane be
2x – y + z = 3 …………………(1)
Let M be the foot of ⊥ drawn from P(2, 3, – 5) on plane (1).
Since D’ ratios of normal to plane (1) are < 2, – 1, 1 >.
∴ eqn. of line PM be
\(\frac{x-2}{2}=\frac{y-3}{-1}=\frac{z+4}{1}\) = t ……………….(2) (say)

So any pointon line (2) be Q (2t + 2, – t + 3, t – 4).
Now point Q be the image of point P if M be the mid-point of tine segment PQ.
∴ Coordinates of point M are \(\left(\frac{2 t+2+2}{2}, \frac{-t+3+3}{2}, \frac{t-4-4}{2}\right)\)
i.e. M \(\left(\frac{2 t+4}{2}, \frac{-t+6}{2}, \frac{t-8}{2}\right)\)
Since point M lies on plane (1).
∴ \(2\left(\frac{2 t+4}{2}\right)-\left(\frac{-t+6}{2}\right)+\left(\frac{t-8}{2}\right)\) = 3
⇒ 4t + 8 + t – 6 + t – 8 = 6
⇒ 6t = 12
⇒ t = 2
∴ Coordinates of Q are (4 + 2, – 2 + 3, 2 – 4) i.e. (6, 1, – 2)
Hence the required image of point P be
Q(6, 1, -2)
A A A
whose position vector be 6 ¡ + j- 2k.

Question 26.
Find the equation of the plane through the intersection of the planes \(\vec{r} \cdot(\hat{i}+3 \hat{j}-\hat{k})\) = 5 and \(\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})\) = 3 and passing through the point (2, 1, – 2). Also find the length of perpendicular from origin to the plane.
Solution:
Given equations of planes are
\(\vec{r} \cdot(\hat{i}+3 \hat{j}-\hat{k})\) = 5
and \(\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})\) = 3
∴ Cartesian eqns. of planes are
x + 3y – z = 5 …………………(1)
and 2x – y + z = 3 …………………(2)
Thus the required eqn. of plane through the intersection of both planes (1) and (2) is given by
(x + 3y – z – 5) + (2x – y + z – 3) = 0 ………………..(3)
Since the plane (3) passes through the point (2, 1, – 2).
(2 + 3 + 2 – 5) + λ (4 – 1 – 2 – 3) = 0
⇒ 2 – 2λ = 0
⇒ λ = 1
putting the value of in eqn. (3) ; we have
(x + 3y – z – 5) + λ (2x – y + z – 3) = 0
⇒ 3x + 2y + 0z – 8 = 0 ……………………(4)
which is the required eqn. of plane.
∴ required length of L from (0, 0, 0) to plane (4) = \(\frac{|3(0)+2(0)+0-8|}{\sqrt{3^2+2^2}}\)
= \(\frac{8}{\sqrt{13}}\) units.

Question 27.
Prove that the line containing the points A (0, 4, 3), B (- 1, – 5, – 3) intersects the line containing the points C (- 2, – 2, 1). L) (1, 1, – 1). Also find the equation of the plane in which they lie.
Solution:
The equation of line through the points A (0, 4, 3) and B (- 1, – 5, – 3) is given by \(\frac{x-0}{-1-0}=\frac{y-4}{-5-4}=\frac{z-3}{-3-3}\)
i.e. \(\frac{x}{-1}=\frac{y-4}{-9}=\frac{z-3}{-6}\)
i.e. \(\frac{x}{1}=\frac{y-4}{9}=\frac{z-3}{6}\) ………………(1)
and eqn. of line through the points C (- 2, – 2, 1) and D (1, 1, – 1) is given by
\(\frac{x+2}{1-2}=\frac{y+2}{1+2}=\frac{z-1}{-1-1}\)
i.e. \(\frac{x+2}{3}=\frac{y+2}{3}=\frac{z-1}{-2}\) ……………………(2)
For intersection of line (1) and line (2). So any point online (1)
i.e. (t, 9t + 4, 6t + 3) lies on line (2).
∴ \(\frac{t+2}{3}=\frac{9 t+4+2}{3}=\frac{6 t+3-1}{-2}\) are consistent.
iff \(\frac{t+2}{3}=\frac{9 t+6}{3}\) and \(\frac{9 t+6}{3}=\frac{6 t+2}{-2}\) are consistent
iff 8t = – 4 and – 18t – 12 = 18t + 6 are consistent
iff t = – \(\frac{1}{2}\) and t = – \(\frac{1}{2}\) , which is true
∴ line (1) and line (2) are intersects.
The eqn. of plane containing the line (1) is given by
a (x – 0) + b (y – 4) + c (z – 3) = 0
where a + 9b + 6c = 0 ………………….(4)
Also plane (3) containing line (2).
∴ 3a + 3b – 2c = 0 ……………….(5)
On solving eqn. (4) and eqn. (5) ; we have
\(\frac{a}{-18-18}=\frac{b}{18+2}=\frac{c}{3-27}\)
i.e. \(\frac{a}{-36}=\frac{b}{20}=\frac{c}{-24}\)
i.e. \(\frac{a}{9}=\frac{b}{-5}=\frac{c}{6}\) = k (say), k ≠ 0
∴ a = 9k ;
b = – 5k ;
c = 6k
putting the value of a, b, c in eqn. (3) ; we have
9kx – 5k (y – 4) + 6k (z – 3) = 0
⇒ 9x – 5y + 6z + 2 = 0 [∵ k ≠ 0]
which is the required eqn. of plane.

Question 28.
Find the vector equation of the plane passing through the point (1, – 1, – 1) and perpendicularto each of the planes \(\vec{r} \cdot(\hat{i}-2 \hat{j}-8 \hat{k})\) = 0 and \(\vec{r} \cdot(2 \hat{i}+5 \hat{j}-\hat{k})\) + 11 = 0.
Solution:
The vector eqns. of given planes are
\(\vec{r} \cdot(\hat{i}-2 \hat{j}-8 \hat{k})\) = 0
and \(\vec{r} \cdot(2 \hat{i}+5 \hat{j}-\hat{k})\) + 11 = 0
∴ cartesian eqns. of given planes are ;
x – 2y – 8z = 0 ……………..(1)
and 2x + 5y – z + 11 = 0 …………………..(2)
Let the equation of plane through the point (1, – 1, – 1) is given by
a (x – 1) + b (y + 1) + c (z + 1) = 0
where < a, b, c > are the direction ratios of the normal to plane (3).
Since plane (3) is ⊥ to plane (1) and (2).
∴ a – 2b – 8c = 0 ………………..(4)
and 2a + 5b – c = 0 ………………..(5)
On solving eqn. (4) and eqn. (5) ; we have
\(\frac{a}{2+40}=\frac{b}{-16+1}=\frac{c}{5+4}\)
i.e. \(\frac{a}{42}=\frac{b}{-15}=\frac{c}{9}\)
i.e. \(\frac{a}{14}=\frac{b}{-5}=\frac{c}{3}\) = k (say) ; k ≠ 0
∴ a = 14 k ;
b = – 5k ;
c = 3k
putting the values of a, b, c in eqn. (3) ; we have
14k (x – 1) – 5k (y + 1) + 3k (z + 1) = 0
⇒ 14x – 5y + 3z – 16 = 0 [∵ k ≠ 0] ………………(6)
which is the required eqn. of plane.
If \(\vec{r}\) be the P.V of point P(x, y, z).
Then \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\)
Then eqn. (6) becomes ;
\(\vec{r} \cdot(14 \hat{i}-5 \hat{j}+3 \hat{k})\) = 16, gives the vector eqn. of plane.

Question 29.
If the planes \(\vec{r} \cdot(2 \hat{i}-\lambda \hat{j}+\hat{k})\) = 3 and \(\vec{r} \cdot(4 \hat{i}+\hat{j}-\mu \hat{k})\) = 5 are parallel, find λ and µ.
Solution:
Given planes in cartesian form are
2x – y + z = 3 …………………..(1)
and 4x – y – 4z = 5 ………………….(2)
Since planes (1) and (2) are parallel.
∴ D’ ratios of normal to planes (1) and (2) are proportional.
i.e. \(\frac{2}{4}=-\frac{\lambda}{1}=\frac{1}{-\mu}\)
Thus, – λ = \(\frac{1}{2}\)
⇒ λ = – \(\frac{1}{2}\)
and – µ = 2
⇒ µ = – 2.

Question 30.
Find the distance between the planes \(\vec{r} \cdot(\hat{i}+\hat{j}-\hat{k})\) + 4 = 0 and \(\vec{r} \cdot(\hat{i}+\hat{j}-\hat{k})\) + 5 = 0.
Solution:
Given planes in cartesian form are
x + y – z + 4 = 0 ………………….(1)
x + y – z + 5 = 0 ………………….(2)
Since the D’ ratios of normal to both planes (1) and (2) are proportional.
∴ planes (1) and (2) are parallel.
∴ Distance between parallel planes distance of any point on plane (1) from plane (2).
To find point on plane (1) we put y = z = 0 in eqn. (1) ; we have
∴ (- 4, 0, 0) be a point on plane (1).
∴ required distance between parallel planes distance of (- 4, 0, 0) from plane (2)
= \(\frac{|-4+0-0+5|}{\sqrt{1^2+1^2+(-1)^2}}\)
= \(\frac{1}{\sqrt{3}}\) units