JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Three Dimensional Geometry

The plane passing through the point

(4, - 1, 2)

and parallel to the lines

\frac{x + 2}{3} = \frac{y - 2}{- 1} = \frac{z + 1}{2}

and

\frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 4}{3}

also passes through the point:

(A) $(1, 1, - 1)$
(B) $(1, 1, 1)$ ✓ Correct
(C) $(- 1, - 1, - 1)$
(D) $(- 1, - 1, 1)$

Answer: (B)

Solution

The two lines have direction vectors

d_{1} = (3, - 1, 2)

and

d_{2} = (1, 2, 3)

. Since the plane is parallel to both, its normal is

n = d_{1} \times d_{2} = (- 7, - 7, 7)

, which we may take as

(1, 1, - 1)

. The plane through

(4, - 1, 2)

1 (x - 4) + 1 (y + 1) - 1 (z - 2) = 0

, i.e.

x + y - z = 1

. Substituting the point

(1, 1, 1)

gives

1 + 1 - 1 = 1

, so it lies on this plane.

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Single correct option

Practise more: all JEE Main PYQs on Three Dimensional Geometry →

More JEE Main questions on Three Dimensional Geometry

Let $(α, β, γ)$ be the co-ordinates of the foot of the perpendicular drawn from the point $(5, 4, 2)$ on the line $r = (- \hat{i} + 3 \hat{j} + \hat{k}) + λ (2 \hat{i} + 3 \hat{j} - \hat{k})$ .
Then the length of the projection of the vector …
JEE Main 2026View solution →
Let the line L pass through the point $(- 3, 5, 2)$ and make equal angles with the positive coordinate axes. If the distance of L from the point $(- 2, r, 1)$ is $\frac{14}{3}$ , then the sum of all possible values of $r$ is :
JEE Main 2026View solution →
Let the line $L_{1}$ be parallel to the vector $- 3 \hat{i} + 2 \hat{j} + 4 \hat{k}$ and pass through the point $(2, 6, 7)$ and the line $L_{2}$ be parallel to the vector $2 \hat{i} + \hat{j} + 3 \hat{k}$ and pass through the point $(4, 3, 5)$ . If the line $L_{3}$ is parallel to the vector …
JEE Main 2026View solution →
Let $P (α, β, γ)$ be the point on the line
$\frac{x - 1}{2} = \frac{y + 1}{- 3} = z$ at a distance $414$ from the point $(1, - 1, 0)$ and nearer to the origin. Then the shortest distance, between the lines …
JEE Main 2026View solution →
If the image of the point $P (1, 2, a)$ in the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{7 - z}{2}$ is $Q (5, b, c)$ , then $a^{2} + b^{2} + c^{2}$ is equal to
JEE Main 2026View solution →
Let L be the line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z + 3}{6}$ and let S be the set of all points $(a, b, c)$ on L, whose distance from the line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z - 9}{0}$ along the line L is 7. Then $\sum_{(a, b, c) \in S} (a + b + c)$ is equal to :
JEE Main 2026View solution →