The plane containing the line (x - 1/1) = (y - 2/2) = (z - 3/3) and parallel to the line (x/1) = (y/1) = (z/4) passes through the point :

Question

The plane containing the line (x - 1/1) = (y - 2/2) = (z - 3/3) and parallel to the line (x/1) = (y/1) = (z/4) passes through the point : (A) (1, - 2, 5) (B) (1, 0, 5) (C) (0, 3, - 5) (D) (- 1, - 3, 0)

Tardigrade Admin · Accepted Answer

Normal vector = |i&j&k 1&2&3 1&1&4| = 5i- hatj - hatk point (1,2,3) lies in plane so equation of plane = 5 (x - 1) - 1 (y - 2) - 1 (z - 3) = 0 5x - y - z = 0 so option [2] is correct

Q. The plane containing the line $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}$ and parallel to the line $\frac{x}{1} = \frac{y}{1} = \frac{z}{4}$ passes through the point :

(1, - 2, 5)

(1, 0, 5)

(0, 3, - 5)

(- 1, - 3, 0)

Normal vector $= ∣ ∣ i 11 j 21 k 34 ∣ ∣ = 5 i - \hat{j} - \hat{k}$
point $(1, 2, 3)$ lies in plane so equation of
plane $= 5 (x - 1) - 1 (y - 2) - 1 (z - 3) = 0$
$5 x - y - z = 0$
so option $[2]$ is correct

Q. The plane containing the line 1x−1​=2y−2​=3z−3​ and parallel to the line 1x​=1y​=4z​ passes through the point :