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Question 1304a
Vectors II: Lines and Planes
Equation of a Plane from the Normal Vector
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The line
l
{l}
l
and plane
p
{p}
p
are given by
l
:
r
=
(
2
2
−
5
)
+
λ
(
2
−
2
−
4
)
,
λ
∈
R
,
p
:
r
⋅
(
−
4
4
−
5
)
=
13.
\begin{align*} &l: \mathbf{r} = \begin{pmatrix} 2 \\ 2 \\ - 5 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ - 2 \\ - 4 \end{pmatrix}, \; \lambda \in \mathbb{R}, \\ &p: \mathbf{r} \cdot \begin{pmatrix} - 4 \\ 4 \\ - 5 \end{pmatrix} = 13. \end{align*}
l
:
r
=
2
2
−
5
+
λ
2
−
2
−
4
,
λ
∈
R
,
p
:
r
⋅
−
4
4
−
5
=
13.
Find the acute angle,
θ
,
{\theta,}
θ
,
between the the line and the plane.
Attempt
θ
=
{\theta = }
θ
=
∘
{^\circ}
∘
Answer
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