Vector: Angle

**Simplicity** · March 6th 2008, 06:22 AM

I don't get the answer required. Where am I wrong? Help would be appreciated.

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Q:

A line has equation $\bold r = 3 \bold i - 5 \bold j + 2 \bold k + \lambda (2 \bold i -4 \bold j + \bold k)$ and a plane has equation $\bold r . (3 \bold i - \bold j - 5 \bold k) = 1$ . Find the acute angle between the line and the plane.

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My Method:

$\bold n . \bold n = | \bold n | | \bold b | \cos \phi$
$\begin{pmatrix} 3 \\ -1 \\ 5 \end{pmatrix} . \begin{pmatrix} 2 \\ -4 \\ 1 \end{pmatrix} = (\sqrt {35}) (\sqrt {21}) \cos \phi$
$\therefore \cos \phi = \frac {15}{7 \sqrt {15}} = 56.4^{\circ}$
$\implies \theta = 33.6^{\circ} (90^{\circ} - 56.4^{\circ})$ .

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The Correct Answer:

$10.6^{\circ}$ .

**TheEmptySet** · March 6th 2008, 06:37 AM

To find the angle between a vector and a plane you need to find the angle between their normal vectors.

<2,-4,1> is normal to the line and
<3,-1,-5> is normal to the plane

so this gives..

$5=\sqrt{21}\sqrt{35}\cos(\theta)$

solving for theta gives...

$\theta=79.35$

so the acute angle is

$\theta=90-79.35=10.65$

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