Interactive ISC Class 12 Maths OP Malhotra Solutions Chapter 24 The Plane Ex 24(d) engage students in active learning and exploration.
S Chand Class 12 ICSE Maths Solutions Chapter 24 The Plane Ex 24(d)
Question 1.
Find.the equation of the planes through the intersection of the planes
x + 2 y + 3 z + 5 = 0,2 x – 4 y + z – 3 = 0 and the point (0, 1, 0)
Answer:
Given equations of planes are :
x + 2 y + 3 z + 5 = 0
2 x – 4 y + z – 3 = 0
and Thus the eqn. of plane through the inter-section of planes (1) and (2) is given by
(x + 2 y + 3 z + 5) + λ(2 x – 4 y + z – 3) = 0
Also plane given by eqn. (3) passes through the point (0, 1, 0).
∴(0 + 2 + 0 + 5) + λ(0 – 4 + 0 – 3) = 0
⇒ 7 – 7 λ = 0
⇒ λ = 1
Putting the value of λ in eqn. (3); we have
(x + 2 y + 3 z + 5) + (2 x – 4 y + z – 3) = 0
⇒ 3 x – 2 y + 4 z + 2 = 0
which is the required eqn. of plane.
Question 2.
Find the equation of the plane containing the line of intersection of the plane x + y + z – 6 = 0 and 2 x + 3 y + 4 z + 5 = 0 passing through the point (1, 1, 1).
Answer:
The eqn. of plane passing through the line of intersection of given planes be
\(\vec{r}\) (\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)) = 6 and \(\vec{r}\) (2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\)) + 5 = 0
be given by [\(\vec{r}\) (\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)) – 6] + λ[\(\vec{r}\) (2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\)) + 5] = 0
⇒ \(\vec{r}\) [(1 + 2 λ) \(\hat{i}\) + (1 + 3 λ) \(\hat{j}\) + (1 + 4 λ) \(\hat{k}\)] – 6 + 5 λ = 0
Since the plane (1) passes through the point (1, 1, 1)
∴ The vector \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\) lies on eqn ………… (1)
we get (\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\))
[ (1 + 2 λ) \(\hat{i}\) + (1 + 3 λ) \(\hat{j}\) + (1 + 4 λ) \(\hat{k}\)] – 6 + 5 λ = 0
⇒ 1(1 + 2 λ) + 1(1 + 3 λ) + 1(1 + 4 λ) – 6 + 5 λ = 0
⇒ 14 λ = 3 ⇒ λ = \(\frac{3}{14}\)
putting the value of λ in eqn. (1)
we have \(\vec{r}\) [\(\frac{20}{14}\) \(\hat{i}\) + \(\frac{23}{14}\) \(\hat{j}\) + \(\frac{26}{14}\) \(\hat{k}\)] – \(\frac{69}{14}\) = 0
⇒ \(\vec{r}\) [(20 \(\hat{i}\) + 23 \(\hat{j}\) + 26 \(\hat{k}\)] = 69 be the required eqn. of plane.
Question 3.
Find the direction ratios of the normal to the plane passing through the point (2, 1, 3) and the line of intersection of the planes x + 2 y + z = 3 and 2 x – y – z = 5.
Answer:
The given planes are
x + 2 y + z = 3
2 x – y – z = 5
and
Thus, the eqn. of plane through the line of intersection of planes (1) and (2) is given by
(x + 2 y + z – 3) + λ(2 x – y – z – 5) = 0
Now plane (3) passes through the point (2, 1, 3).
∴ (2 + 2 + 3 – 3) + λ(4 – 1 – 3 – 5) = 0
⇒ 4 + (-5 λ) = 0 ⇒ λ = 4 / 5
Putting the value of λ = 4 / 5 in eqn. (3); we have
(x + 2 y + z – 3) + \(\frac{4}{5}\) (2 x – y – z – 5) = 0 ⇒ 13 x + 6 y + z – 35 = 0
Thus the direction ratios of normal to plane (4) are < 13, 6, -1 >
Question 4.
Find the equation of the plane passing through the intersection of the planes
\(\vec{r}\) (2 \(\hat{i}\) + \(\hat{j}\) + 3 \(\hat{k}\)) = 7 and \(\vec{r}\) (2 \(\hat{i}\) + 5 \(\hat{j}\) + 3 \(\hat{k}\)) = 9 the point (2, 1, 3)
Answer:
The eqn. of plane passing through the line of intersection of given planes be
\(\vec{r}\) (2 \(\hat{i}\) + \(\hat{j}\) + 3 \(\hat{k}\)) = 7 and \(\vec{r}\) (2 \(\hat{i}\) + 5 \(\hat{j}\) + 3 \(\hat{k}\)) = 9
is given by [\(\vec{r}\) (2 \(\hat{i}\) + \(\hat{j}\) + 3 \(\hat{k}\)) – 7] + λ[\(\vec{r}\) (2 \(\hat{i}\) + 5 \(\hat{j}\) + 3 \(\hat{k}\)) – 9] = 0
⇒ \(\vec{r}\) [(2 + 2 λ) \(\hat{i}\) + (1 + 5 λ) \(\hat{j}\) + (3 + 3 λ) \(\hat{k}\)] – 7 – 9 λ = 0
Since the plane (1) passes through the point (2, 1, 2).
∴(2 \(\hat{i}\) + \(\hat{j}\) + 3 \(\hat{k}\)) [(2 + 2 λ) \(\hat{i}\) + (1 + 5 λ) \(\hat{j}\) + (3 + 3 λ) \(\hat{k}\)] – 7 – 9 λ = 0
⇒ 2(2 + 2 λ) + (1 + 5 λ) + 3(3 + 3 λ) – 7 – 9 λ = 0 ⇒ 9 λ = -7 ⇒ λ = \(-\frac{7}{9}\)
putting the value of λ in eqn. (1); we get
\(\vec{r}\) [\(\frac{4}{9}\) \(\hat{i}\) – \(\frac{26}{9}\) \(\hat{j}\) + \(\frac{6}{9}\) \(\hat{k}\)] – 7 + 7 = 0
⇒ \(\vec{r}\) (2 \(\hat{i}\) – 13 \(\hat{j}\) + 3 \(\hat{k}\)) = 0 be the required eqn. of plane.
Question 5.
Find the equation of the plane passing through the intersection of the planes 2 x + 3 y – z + 1 = 0, x + y – 2 z + 3 = 0 and perpendicular to the plane 3 x – y – 2 z – 4 = 0.
Answer:
The eqn. of plane through the intersection of two given plane
2 x + 3 y – z + 1 = 0 and x + y – 2 z + 3 = 0 be given by :
2 x + 3 y – z + 1 + λ(x + y – 2 z + 3) = 0
⇒ (2 + λ) x + (3 + λ) y + (-1 – 2 λ) z + 1 + 3 λ = 0
∴ Direction ratios of normal to plane (1) are < 2 + λ, 3 + λ, -1 – 2 λ >
Since the plane (1) is ⊥ to given plane 3 x – y – 2 z – 4 = 0
where d ratios of normal to plane are < 3, -1, -2 >
∴(2 + λ) 3 + (3 + λ)(-1) + (-1 – 2 λ)(-2) = 0 ⇒ 6 λ + 5 = 0 ⇒ \(-\frac{5}{6}\)
putting the value of λ in eqn. (1); we have
(2 – \(\frac{5}{6}\)) x + (3 – \(\frac{5}{6}\)) y + (-1 + \(\frac{5}{3}\)) z + 1 – \(\frac{5}{2}\) = 0
⇒ \(\frac{7}{6}\) x + \(\frac{13}{6}\) y + \(\frac{2}{3}\) z – \(\frac{3}{2}\) = 0
⇒ 7 x + 13 y + 4 z – 9 = 0 which is the required eqn. of plane.
Question 6.
Find the equation of the plane through the point 2 \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\) and passing through the line of intersection of the planes \(\vec{r}\) (\(\hat{i}\) + 3 \(\hat{j}\) – \(\hat{k}\)) = 0 and \(\vec{r}\) (\(\hat{j}\) + 2 \(\hat{k}\)) = 0.
Answer:
The eqn. of given planes are :
\(\vec{r}\) (\(\hat{i}\) + 3 \(\hat{j}\) – \(\hat{k}\)) = 0
\(\vec{r}\) (\(\hat{j}\) + 2 \(\hat{k}\)) = 0
and
Thus the eqn. of plane through the line of intersection of planes (1) and (2) is given by
\(\vec{r}\) (\(\hat{i}\) + 3 \(\hat{j}\) – \(\hat{k}\)) + λ\(\vec{r}\) (\(\hat{j}\) + 2 \(\hat{k}\)) = 0
⇒ \(\vec{r}\) [\(\hat{i}\) + (3 + λ) \(\hat{j}\) + (-1 + 2 λ) \(\hat{k}\)] = 0
Since the plane (3) passes through the point where P.V is 2 \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)
∴ (2 \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)) [\(\hat{i}\) + (3 + λ) \(\hat{i}\) + (-1 + 2 λ) \(\hat{k}\)] = 0
⇒ 2(1) + 1(3 + λ) – 1(-1 + 2 λ) = 0
⇒ 2 + 3 + λ + 1 – 2 λ = 0 ⇒ 6 – λ = 0 ⇒ λ = 6
putting the value of λ = 6 in eqn. (3); we get
\(\vec{r}\) (\(\hat{i}\) + 9 \(\hat{j}\) + 11 \(\hat{k}\)) = 0 which is the required eqn. of plane.
Question 7.
Find the vector equation of the plane through the point (1, 4, -2) and perpendicular to the line of intersection of the planes x + y + z = 10,2 x – y + 3 z = 18.
Answer:
The equations of given planes are
and
x+y+z-10=0
2 x-y+3 z-18=0
∴ d ratios of normal to planes (1) and (2) are < 1, 1, 1> and < 2, -1, 3 >.
Let the eqn. of plasses through the point (1,4,-2) is given by
a(x – 1) + b(y – 4) + c(z + 2) = 0
where < a, b, c> are the direction ratios of normal to plane.
Since plane (3) is ⊥ to plane (1) and (2)
∴ a + b + c = 0
2 a – b + 3 c = 0
Solving eqn. (4) and eqn. (5) by cross multiplication method, we have
\(\frac{a}{3+1}\) = \(\frac{b}{2-3}\) = \(\frac{c}{-1-2}\)
i.e. \(\frac{a}{4}\) = \(\frac{b}{-1}\) = \(\frac{c}{-3}\) = k (say)
i.e. a = 4 k ; b = -k ; c = -3 k where k ≠ 0
4 k(x – 1) – k(y – 4) – 3 k(z + 2) = 0
⇒ 4 x – y – 3 z = 6 which is the reqd. eqn. of plane
Question 8.
Find the equation of the plane through the intersection of the planes x + 3 y + 6 = 0 and 3 x – y – 4 z = 0 and whose perpendicular distance from the origin is unity.
Answer:
The eqns. of given planes are
x + 3 y + 6 = 0
3 x – y – 4 z = 0
and
Thus the eqn. of plane through the line of intersection of planes (1) and (2) be given by
(x + 3 y + 6) + λ(3 x – y – 4 z) = 0
⇒ (1 + 3 λ) x + (3 – λ) y – 4 λ z + 6 = 0
It is given that ⊥ distance from origin (0, 0, 0) to plane (3) = 1
⇒ \(\frac{|(1+3 λ) 0+(3-λ) 0-4 λ \times 0+6|}{\sqrt{(1+3 λ)^2+(3-λ)^2+(-4 λ)^2}}\) = 1
⇒ 6 = \(\sqrt{(1+3 λ)^2+(3-λ)^2+16 λ^2}\)
On squaring both sides; we have
36 = (1 + 3 λ)2 + (3 – λ)2 + 16 λ2
⇒ 36 = 1 + 9 λ2 + 6 λ + 9 + λ2 – 6 λ + 16 λ2
⇒ 26 λ2 = 26
⇒ λ2 = 1
⇒ λ = ± 1
Putting the values of λ in eqn. (3); we have
(1 + 3) x + (3 – 1) y – 4 z + 6 = 0 and (1 – 3) x + (3 + 1) y + 4 z + 6 = 0
i.e. 4 x + 2 y – 4 z + 6 = 0 and -2 x + 4 y + 4 z + 6 = 0
i.e. 2 x + y – 2 z + 3 = 0 and x – 2 y – 2 z – 3 = 0
which are the required eqns. of planes.
Question 9.
Find the equation of the plane passing through the intersection of the planes 4 x – y + z = 10 and x + y – z = 4 and parallel to the line with direction ratios 2,1,1. Also, find the perpendicular distance of (1, 1, 1) from this plane.
Answer:
The eqn. of plane through the line of intersection of two given planes
4 x – y + z = 10 and x + y – z = 4 is given by
(4 x – y + z – 10) + λ(x + y – z – 4) = 0
i.e. (4 + λ) x + (-1 + λ) y + (1 – λ) z – 10 – 4 λ = 0
∴ D ratios of normal to plane, are < 4 + λ, -1 + λ, 1 – λ >
Also plane (1) parallel to line having direction ratios < 2, 1, 1 >
Thus normal to plane (1) is ⊥ to given line.
∴ 2(4 + λ) + (-1 + λ) 1 + (1 – λ) 1 = 0
⇒ 8 + 2 λ + λ – 1 + 1 – λ = 0
⇒ 8 + 2 λ = 0 ⇒ λ = -4
Putting the values of λ in eqn. (1) ; we have
-5 y + 5 z + 6 = 0 ⇒ 5 y – 5 z – 6 = 0
∴ required ⊥ distance of point (1, 1, 1) from plane (2)
= \(\frac{|5 \times 1-5 \times 1-6|}{\sqrt{5^2+5^2}}\) = \(\frac{6}{\sqrt{50}}\) = \(\frac{6}{.5 \sqrt{2}}\) = \(\frac{3 \sqrt{2}}{5}\) units
Question 10.
Find the equation of the plane which passes through the line of intersection of the planes x + 5 y – 2 z = 6 and 5 x – 4 y + 5 z = 2 and parallel to the line joining the points (5, 1, 4) and (4, -2, 3).
Answer:
The eqn. of plane through the line of intersection of given planes
x + 5 y – 2 z – 6 = 0 and 5 x – 4 y + 5 z – 2 = 0 be given by
(x + 5 y – 2 z – 6) + λ(5 x – 4 y + 5 z – 2) = 0
i.e. (1 + 5 λ) x + (5 – 4 λ) y + (-2 + 5 λ) z + (-6 – 2 λ) = 0
∴ D ratios of normal to plane (1) are < 1 + 5 λ, 5 – 4 λ,-2 + 5 λ >
Since the plane (1) is || to line joining (5, 1, 4) and (4, -2, 3)
∴D ratios of given line are < 4 – 5, -2 – 1, 3 – 4 > i.e. < -1, -3, -1 > i.e. < 1, 3, 1 > Thus, normal to plane (1) is ⊥ to given line.
(1 + 5 λ) 1 + (5 – 4 λ) 3 + (-2 + 5 λ) 1 = 0
5 λ + 1 + 15 – 12 λ – 2 + 5 λ = 0
⇒ -2 λ + 14 = 0
⇒ λ = 7
⇒ 5 λ + 1 + 15 – 12 λ – 2 + 5 λ = 0
⇒ -2 λ + 14 = 0 ⇒ λ = 7
Putting the value of λ in eqn. (1); we get
36 x – 23 y + 33 z – 20 = 0 which is the reqd. eqn of plane.
Question 11.
Find the equation to the plane through the line of intersection of the planes a x + b y + c z + d = 0, a’ x + b’ y + c’ z + d’ = 0 and
(i) parallel to the x-axis,
(ii) perpendicular to x y-plane.
Answer:
(i) The eqn. of plane through the line of intersection of given planes
a x + b y + c z + d = 0 and a’ x + b’ y + c’ z + d’ = 0 is given by
(a x + b y + c z + d) + λ(a’ x + b’ y + c’ z + d’) = 0
⇒ (a + λ a’) x + (b + λ b’) y + (c + λ c’) z + d + λ d’ = 0
Now plane (1) is || to x-axis
∴ perpendicular to Y O Z plane i.e. x = 0 ⇒ 1 x + 0 y + 0 z = 0
Thus, (a + λ a’) 1 + (b + λ b’) 0 + (c + λ c’) 0 = 0
⇒ λ = \(\frac{-a}{a’}\)
Putting the value of λ in eqn. (1); we have
(b – \(\frac{a b’}{a’}\)) y + (c – \(\frac{a c’}{a’}\)) z + d + (\(\frac{-a}{a’}\)) d’ = 0
⇒ a'(b y + c z + d) = a(b’ y + c’ z + d’) be the reqd. plane.
(ii) Now plane (1) is ⊥ to plane x y i.e. z = 0 i.e. 0 x + 0 y + 1 z = 0
So normal to plane (1) is ⊥ to plane (3).
∴(a + λ a’) 0 + (b + λ b’) 0 + (c + λ c’) 1 = 0
⇒ λ = \(\frac{-c}{c’}\)
putting the value of λ in eqn. (1); we have
(a – \(\frac{a’ c}{c’}\)) x + (b – \(\frac{b’ c}{c’}\)) y + d + (\(\frac{-c}{c’}\)) d’ = 0
⇒ c'(a x + b y + d) = c(a’ x + b’ y + d’)
which is the reqd. eqn. of plane.
Question 12.
Find the equation of the plane which passes through the y-axis and the point (4, 2, -3).
Answer:
The eqn. of plane passes through the y-axis i.c. intersection of x y plane and y z plane i.e. through the planes z = 0 and x = 0 be given by
z + k x = 0
Further eqn. (1) passes through the point (4, 2, -3).
∴-3 + 4 k = 0 ⇒ k = \(\frac{3}{4}\)
putting the value of k in eqn. (1) ; we get
z + \(\frac{3}{4}\) x = 0
⇒ 3 x + 4 z = 0 be the required plane.
Question 13.
Find the equation of the plane passing through the line of intersection of the planes x + 2 y + 3 z – 5 = 0 and 3 x – 2 y – z + 1 = 0 and cutting off equal intercepts on x-axis and z-axis.
Answer:
The eqn. of piane through the line of intersection of given plane
x + 2 y + 3 z – 5 = 0 and 3 x – 2 y – z + 1 = 0 be given by
(x + 2 y + 3 z – 5) + λ(3 x – 2 y – z + 1) = 0
i.e. (1 + 3 λ) x + (2 – 2 λ) y + (3 – λ) z = 5 – λ
⇒ \( \frac{(1+3 λ) x}{5-λ}\) + \(\frac{(2-2 λ) y}{5-λ}\) + \(\frac{(3-λ) z}{5-λ}\) = 1
⇒ \(\frac{x}{(5-λ)}\) + \(\frac{x}{1+3 λ}\) + \(\frac{5-λ}{2-2 λ}\) + \(\frac{5-λ}{3-λ}\) = 1
Thus the intercepts cut off by plane (2) on x-axis and z-axis are \(\frac{5-λ}{1+3 λ}\) and \(\frac{5-λ}{3-λ}\).
According to given condition ; we have
\(\frac{5-λ}{1+3 λ}\)
= \(\frac{5-λ}{3-λ}\)
⇒ (5 – λ)[3 – λ – 1 – 3 λ] = 0
⇒ (5 – λ)(2 – 4 λ) = 0
⇒ λ = 5, \(\frac{1}{2}\) ; since λ ≠ 5
∴ λ = \(\frac{1}{2}\)
putting the value of λ in eqn. (1); we have
(x + 2 y + 3 z – 5) + \(\frac{1}{2}(3 x-2 y-z+1)\) = 0
⇒ 5 x + 2 y + 5 z – 9 = 0 which is the required plane.
Question 14.
Find the vector equation of the plane containing the line of intersection of the planes x – 3 y + 2 z – 5 = 0 and 2 x – y + 3 z – 1 = 0 and passing through the point (1, -2, 3).
Answer:
The equation of any plane passing through the line of intersection of given planes
x – 3 y + 2 z – 5 = 0 and 2 x – y + 3 z – 1 = 0
(x – 3 y + 2 z – 5) + λ(2 x – y + 3 z – 1) = 0
⇒ (1 + 2 λ) x + (-3 – λ) y + (2 + 3 λ) z – 5 – λ = 0
Since plane (1) passing through the point (1, -2, 3)
∴ (1 + 2 λ) 1 + (-3 – λ)(-2) + (2 + 3 λ) 3 – 5 – λ = 0
⇒ 1 + 2 λ + 6 + 2 λ + 6 + 9 λ – 5 – λ = 0 ⇒ 12 λ = -8 ⇒ λ = \(-\frac{2}{3}\)
Since plane (1) passing through the point (1, -2, 3)
∴ (1 + 2 λ) 1 + (-3 – λ)(-2) + (2 + 3 λ) 3 – 5 – λ = 0
⇒ 1 + 2 λ + 6 + 2 λ + 6 + 9 λ – 5 – λ = 0 ⇒ 12 λ = -8 ⇒ λ = \(-\frac{2}{3}\)
putting the value of λ in eqn. (1); we get
\(-\frac{x}{3}\) – \(\frac{7}{3}\) y + 0 z – \(\frac{13}{3}\) = 0
⇒ x + 7 y + 13 = 0
⇒ (x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\)) (\(\hat{i}\) + 7 \(\hat{j}\) + 0 \(\hat{k}\)) + 13 = 0
⇒ \(\vec{r}\) (\(\hat{i}\) + 7 \(\hat{j}\) + 0 \(\hat{k}\)) + 13 = 0
be the required vector eqn. of plane in scalar product form.
Question 15.
Find the equation of the plane passing through the line of intersection of the pianes 2 x + y – z = 3 and 5 x – 3 y + 4 z + 9 = 0 and parallel to the line \(\frac{x-1}{2}\) = \(\frac{y-3}{4}\) = \(\frac{z-5}{5}\)
Answer:
The eqns. of given planes are ;
2 x + y – z = 3
5 x – 3 y + 4 z + 9 = 0
Thus, the eqn. of plane passing through the line of intersection of planes (1) and (2) be given by
(2 x + y – z – 3) + λ(5 x – 3 y + 4 z + 9) = 0
(2 + 5 λ) x + (1 – 3 λ) y + (-1 + 4 λ) z – 3 + 9 λ = 0
∴D’ ratios of normal to plane are < 2 + 5 λ, 1 – 3 λ, -1 + 4 λ >
The eqn. of given line be \(\frac{x-1}{2}\) = \(\frac{y-3}{4}\) = \(\frac{z-5}{5}\)
Sine plane (3) is || to line (4).
Thus normal to plane (3) is ⊥ to line (4).
(2 + 5 λ) 2 + (1 – 3 λ) 4 + (-1 + 4 λ) 5 = 0
⇒ 4 + 10 λ + 4 – 12 λ – 5 + 20 λ = 0
⇒ 18 λ + 3 = 0
⇒ x = -1 / 6
putting the value of λ in eqn. (3); we have
(2 x + y – z – 3) – \(\frac{1}{6}(5 x-3 y+4 z+9)\) = 0
⇒ 7 x + 9 y – 10z – 27 = 0 which is the required plane.
Question 16.
Show that the lines \(\frac{x+4}{3}\) = \(\frac{y+6}{5}\) = \(\frac{z-1}{-2}\) and 3 x – 2 y + z + 5 = 0 = 2 x + 3 y + 4 z – 4 intersect. Find the equation of the plane in which they lie and also their point of intersection.
Answer:
The eqn. of given line be
\(\frac{x+4}{3}\) = \(\frac{y+6}{5}\) = \(\frac{z-1}{-2}\) = t (say)
any pont on line be given by (3 t – 4, 5 t – 6, -2 t + 1) eqn. of given planes are
3 x – 2 y + z + 5 = 0
2 x + 3 y + 4 z – 4 = 0
and
Now given line (1) intersects given planes if any point on given line (1) lies on planes (2) and (3).
and
3(3 t – 4) – 2(5 t – 6) + (-2 t + 1) + 5 = 0
2(3 t – 4) + 3(5 t – 6) + 4(-2 t + 1) – 4 = 0
if 9 t – 12 – 10 t + 12 – 2 t + 1 + 5 = 0 and 6 t – 8 +1 5 t – 18 – 8 t + 4 – 4 = 0
if -3 t + 6 = 0 and 13 t – 26 = 0 if t = 2 and t = 2
Thus given line and pianes intersects.
So point of intersection is given by (3 × 2 – 4, 5 × 2 – 6, -2 × 2 + 1)
i.e. (2, 4, -3) eqn. of any plane through the line of intersection of planes (2) and (3) be given by
(3 x – 2 y + z + 5) + λ(2 x + 3 y + 4 z – 4) = 0
(3 + 2 λ) x + (-2 + 3 λ) y + (1 + 4 λ) z + 5 – 4 λ = 0
Now line (1) lies on plane (4) ∴normal to plane (4) is ⊥ to line (1).
∴(3 + 2 λ) 3 + (-2 + 3 λ) 5 + (1 + 4 λ)(-2) = 0
⇒ 13 λ – 3 = 0
⇒ λ = \(\frac{3}{13}\)
putting the value of λ in eqn. (4); we have
(3 x – 2 y + z + 5) + \(\frac{3}{13}(2 x+3 y+4 z-4)\) = 0
⇒ 45 x – 17 y + 25 z + 53 = 0
which is the required eqn. of plane.