JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Three Dimensional Geometry

Let

P

be a plane containing the line

\frac{x - 1}{3} = \frac{y + 6}{4} = \frac{z + 5}{2}

and parallel to the line

\frac{x - 3}{4} = \frac{y - 2}{- 3} = \frac{z + 5}{7}

. If the point

(1, - 1, α)

lies on the plane

P

, then the value of

∣5 α ∣

is equal to....

Answer: 38

Solution

The plane contains the point

(1, - 6, - 5)

and is parallel to both

(3, 4, 2)

and

(4, - 3, 7)

, so its normal is

n = (3, 4, 2) \times (4, - 3, 7) = (28 + 6, 8 - 21, - 9 - 16) = (34, - 13, - 25)

The plane is therefore

34 (x - 1) - 13 (y + 6) - 25 (z + 5) = 0

. Substituting

(1, - 1, α)

- 13 (5) - 25 (α + 5) = 0 ⟹ 25 α = - 190 ⟹ α = - \frac{38}{5}

∣5 α ∣ = 38

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Integer answer

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