Write the angle between the line x-1 2 = y-2 1 = z+3 -2 and the plane x + y + 4 = 0.

The given line is parallel to the vector b = i +2 j +2 k and the given plane is normal to the vector n = i + j +0 k . We know that the angle between the line and the plane is given by = b . n | b || n | = ( i +2 j +2 k ) ( i + j +0 k ) | i +2 j +2 k || i + j +0 k | =1+2+0 1+4+41+1+0 =3 32 =1 2 = ^ -1 (1 2 )=45^

Write the angle between the line x-1 2 = y-2 1 = z+3 -2 and the p…

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Question

Write the angle between the line $\frac{\text{x}-1}{2}=\frac{\text{y}-2}{1}=\frac{\text{z}+3}{-\text{2}}$ and the plane x + y + 4 = 0.

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Answer

The given line is parallel to the vector $\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$ and the given plane is normal to the vector $\vec{\text{n}}=\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}$.
We know that the angle $\theta$ between the line and the plane is given by
$\sin\theta=\frac{\vec{\text{b}}.\vec{\text{n}}}{|\vec{\text{b}}||\vec{\text{n}}|}$
$=\frac{\big(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big)\big(\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}\big)}{|\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}||\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}|}$
$=\frac{1+2+0}{\sqrt{1+4+4}\sqrt{1+1+0}}$
$=\frac{3}{3\sqrt{2}}$
$=\frac{1}{\sqrt{2}}$
$\Rightarrow\theta=\sin^{-1}\Big(\frac{1}{\sqrt{2}}\Big)=45^\circ$