Q.Find the angle between the following pair of lines:

$\cos\theta=\dfrac{\vec{d_1}\cdot\vec{d_2}}{|\vec{d_1}||\vec{d_2}|}$ for direction vectors $(3,5,4)$ and $(1,1,2)$.

Use $\cos\theta = \frac{|\vec b_1\cdot\vec b_2|}{|\vec b_1||\vec b_2|}$ for the direction ratios.

The angle between two lines is the angle between their direction vectors, given by $\cos\theta=\frac{\vec d_1\cdot\vec d_2}{|\vec d_1||\vec d_2|}$.

For a line and plane, $\sin\theta=\dfrac{|\vec d\cdot\vec n|}{|\vec d||\vec n|}$; here it equals $\frac{8}{21}$.

For a line and a plane, the sine of the angle equals |direction·normal| / (|direction||normal|).

Angle between line and plane: $\sin\theta = \dfrac{|\vec d\cdot\vec n|}{|\vec d|\,|\vec n|}$.

Angle between a line and a plane uses the line direction and plane normal: $\sin\theta=\frac{|\vec d\cdot\vec n|}{|\vec d||\vec n|}$.

The angle between two lines with direction ratios $d_1,d_2$ satisfies $\cos\theta=\dfrac{d_1\cdot d_2}{|d_1||d_2|}$.

Direction ratios: $d_1=(3,-5,4)$, $d_2=(1,1,2)$.

$d_1\cdot d_2=3-5+8=6$. $|d_1|=\sqrt{9+25+16}=\sqrt{50}=5\sqrt2$, $|d_2|=\sqrt{1+1+4}=\sqrt6$.

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Use $\cos\theta=\frac{\vec b_1\cdot\vec b_2}{|\vec b_1||\vec b_2|}$ on the two direction vectors.

$\vec b_1=(1,2,2)$, $\vec b_2=(3,2,6)$. $\vec b_1\cdot\vec b_2=3+4+12=19$.

$|\vec b_1|=\sqrt{1+4+4}=3$, $|\vec b_2|=\sqrt{9+4+36}=7$.

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The angle between a line (direction $\vec b$) and a plane (normal $\vec n$) satisfies $\sin\theta = \tfrac{|\vec b\cdot\vec n|}{|\vec b||\vec n|}$.

Direction of line: $\vec b = (2,3,6)$; normal to plane: $\vec n = (10,2,-11).$

$\vec b\cdot\vec n = 20+6-66 = -40,\quad |\vec b| = \sqrt{4+9+36} = 7,\quad |\vec n| = \sqrt{100+4+121} = 15.$

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The angle between a line and a plane satisfies $\sin\phi=\frac{|\bar b\cdot\bar n|}{|\bar b|\,|\bar n|}$ (the missing RHS constant of the plane does not affect the angle).

Line direction $\bar b=(2,2,1)$; plane normal $\bar n=(-2,2,-1)$.

$\bar b\cdot\bar n=2(-2)+2(2)+1(-1)=-4+4-1=-1$.

$|\bar b|=\sqrt{4+4+1}=3$, $|\bar n|=\sqrt{4+4+1}=3$.

$\sin\phi=\frac{|-1|}{3\cdot3}=\frac19\Rightarrow \phi=\sin^{-1}\!\left(\frac19\right).$ …