The equation of the plane containing two lines $\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-0}{3}$ and $ \frac{x}{-2}=\frac{y-2}{-3}=\frac{z+1}{-1}$, is
Answer & explanation
Correct answer: option 4
Lines: $\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-0}{3}$ and $ \frac{x}{-2}=\frac{y-2}{-3}=\frac{z+1}{-1}$
vectors parallel to given lines
$\vec{b_1}=2\hat i-\hat j+3\hat k$, $\vec{b_2}=-2\hat i-3\hat j-\hat k$
$\vec n$ (normal to plane) = $\vec{b_1}×\vec{b_2}$
$=\begin{vmatrix}\hat i&\hat j&\hat k\\2&-1&3\\-2&-3&-1\end{vmatrix}=10\hat i-4\hat j-8\hat k$
lines pass through points (1, -1, 0) and (0, 2, -1) considering (1, -1, 0)
we get equation of plane
$10×(x-1)-4×(y+1)-8(z)=0$
$10x-10-4y-4-8z=0$
so $10x-4y-8z=14$
$5x-2y-4z=7$