Let the equations of a line and a plane be $\frac{x + 3}{2} = \frac{y - 4}{3} = \frac{\mathcal{z} + 5}{2}$ and $4x - 2y - \mathcal{z} = 1$, respectively, the

Question

Dr. Abubacker · Accepted Answer

(a)
$4(2) - 2(3) - 1(2) = 0$
Also, point $( - 3,\ 4,\  - 5)$ does not lie on the plane
Therefore, the line is parallel to the plane

Q. Let the equations of a line and a plane be $\frac{x + 3}{2} = \frac{y - 4}{3} = \frac{\mathcal{z} + 5}{2}$ and $4x - 2y - \mathcal{z} = 1$, respectively, the

Explanation

Reviewed by Dr. Abubacker

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