The square of the distance of the point of intersection of the line $\vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(\hat{a}-\hat{j})$, $\mathrm{a} \neq 0$ and $\overrightarrow{\mathrm{r}}=(4 \hat{\mathrm{i}}-\hat{\mathrm{k}})+\mu(2 \hat{\mathrm{i}}+\mathrm{a} \hat{\mathrm{k}})$ from the origin is :

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Three Dimensional Geometry

The square of the distance of the point of intersection of the line

r = (\hat{i} + \hat{j} - \hat{k}) + λ (\overset{a}{^} - \hat{j})

a \neq = 0

and

r = (4 \hat{i} - \hat{k}) + μ (2 \hat{i} + a \hat{k})

from the origin is :

(A)5
(B)10
(C)17✓ Correct
(D)26

Answer: (C)

Solution

Writing the two lines as

(1 + aλ, 1 - λ, - 1)

and

(4 + 2 μ, 0, - 1 + a μ)

and equating components, the

\hat{j}

term gives

1 - λ = 0

, so

λ = 1

; the

\hat{k}

term gives

- 1 = - 1 + a μ

, so

a μ = 0

and, since

a \neq = 0

μ = 0

. The

\hat{i}

term then gives

1 + a = 4

, so

a = 3

. The point of intersection is

(4, 0, - 1)

, whose squared distance from the origin is

4^{2} + 0^{2} + (- 1)^{2} = 17

. The square of the distance of the point of intersection from the origin is

17

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Single correct option

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