Q.Find the angle between line (x−1)/3 = (3−y)/(−1) = (3z+1)/6 and the plane 3x − 5y + 2z = 10.

Write the line in standard form to get its direction ratios, then use $\sin\theta = \dfrac{|\vec b\cdot\vec n|}{|\vec b||\vec n|}$ for a line–plane angle.

First rewrite the line's symmetric form with denominators as direction ratios (coefficient of the variable itself must be $+1$ in the numerator):

$\frac{x-1}{3} = \frac{3-y}{-1} = \frac{3z+1}{6}$

$\dfrac{3-y}{-1} = \dfrac{y-3}{1}$, and $\dfrac{3z+1}{6} = \dfrac{z+1/3}{2}$.

So the line is $\dfrac{x-1}{3} = \dfrac{y-3}{1} = \dfrac{z+1/3}{2}$, giving direction ratios $\vec b = \langle 3,1,2\rangle$.

The plane $3x-5y+2z=10$ has normal $\vec n = \langle 3,-5,2\rangle$.

The angle $\theta$ between the line and the plane satisfies: …

The angle between a line and a plane uses $\sin\theta=\dfrac{|\vec b\cdot\vec n|}{|\vec b||\vec n|}$ with $\vec b$ = line direction, $\vec n$ = plane normal.

Direction of the line through $(2,3,-1)$ and $(1,2,1)$: $\vec b = (1-2,\,2-3,\,1-(-1)) = (-1,-1,2)$.

Normal to the plane $2x+3y-5z=10$: $\vec n=(2,3,-5)$.

Angle between line and plane:

$\sin\theta = \frac{|\vec b\cdot\vec n|}{|\vec b||\vec n|}$

$\vec b\cdot\vec n = (-1)(2)+(-1)(3)+(2)(-5) = -2-3-10=-15$

$|\vec b|=\sqrt{1+1+4}=\sqrt6$, \quad $|\vec n|=\sqrt{4+9+25}=\sqrt{38}$

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Using sinθ = |line direction · plane normal| / (|direction||normal|), the angle works out to sin⁻¹(√(26/69)) ≈ 37.9°.

The line $\dfrac{x-1}{2}=\dfrac{2-y}{7}=\dfrac{3z+6}{12}$ must first be written in standard symmetric form.

$\dfrac{2-y}{7} = \dfrac{-(y-2)}{7} = \dfrac{y-2}{-7}$

$\dfrac{3z+6}{12} = \dfrac{3(z+2)}{12} = \dfrac{z+2}{4}$

So the line is $\dfrac{x-1}{2}=\dfrac{y-2}{-7}=\dfrac{z+2}{4}$, with direction ratios $\vec b=(2,-7,4)$.

The plane $3x+4y-z=8$ has normal $\vec n=(3,4,-1)$.

The angle $\theta$ between the line and the plane satisfies:

$\sin\theta = \dfrac{|\vec b\cdot\vec n|}{|\vec b||\vec n|}$

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