The distance between the line $\vec{r}=2\hat{i}-2\hat{j}+3\hat{k}+\lambda(\hat{i}-\hat{j}+4\hat{k})$ and the plane $\vec{r}\cdot(\hat{i}+5\hat{j}+\hat{k})=5$ is

JEE Main · 2005 · 08 May 2005 · Offline · Three Dimensional Geometry

The distance between the line

r = 2 \hat{i} - 2 \hat{j} + 3 \hat{k} + λ (\hat{i} - \hat{j} + 4 \hat{k})

and the plane

r \cdot (\hat{i} + 5 \hat{j} + \hat{k}) = 5

(A) $\frac{10}{9}$
(B) $\frac{10}{3 3}$ ✓ Correct
(C) $\frac{3}{10}$
(D) $\frac{10}{3}$

Answer: (B)

Solution

The line passes through

(2, - 2, 3)

with direction

(1, - 1, 4)

, and the plane

x + 5 y + z = 5

has normal

(1, 5, 1)

. Since

(1, - 1, 4) \cdot (1, 5, 1) = 1 - 5 + 4 = 0

, the line is parallel to the plane, so the distance between them is the distance from

(2, - 2, 3)

to the plane:

\frac{∣2 + 5 ( - 2 ) + 3 - 5∣}{1 + 25 + 1} = \frac{∣ - 10∣}{27} = \frac{10}{3 3}

JEE Main · 2005 · 08 May 2005 · Offline · 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 →