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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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Find the sine of the angle between the vectors $\overrightarrow{a}=3\hat i+\hat j+2\hat k$ and $\overrightarrow{b}=-2\hat i+2\hat j+4\hat k.$

$\begin{array}{1 1} (A)\;\large\frac{4 \sqrt 5}{\sqrt {21}} \\(B)\;\large\frac{2 \sqrt 5}{21} \\ (C)\;\large\frac{2 \sqrt 5}{\sqrt {21}} \\ (D)\;\large\frac{8 \sqrt 5}{\sqrt {21}}\end{array} $

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  • $ \sin \theta = \large\frac{\overrightarrow a \times \overrightarrow b}{|\overrightarrow a||\overrightarrow b|}$
Let $\overrightarrow a=3\hat i+\hat j+2\hat k$ and $\overrightarrow b=-2\hat i+2\hat j+4\hat k$
We know $|\overrightarrow a \times \overrightarrow b|=|\overrightarrow a||\overrightarrow b| \sin \theta$
$ \sin \theta = \large\frac{\overrightarrow a \times \overrightarrow b}{|\overrightarrow a||\overrightarrow b|}$
Let us find $ = \overrightarrow a \times \overrightarrow b$
$ = \overrightarrow a \times \overrightarrow b=\begin {vmatrix} \hat i & \hat j & \hat k \\ 3 & 1 & 2 \\ -2 & 2 & 4 \end {vmatrix}$
=$\hat i(4-4)-\hat j(12+4)+\hat k(6+2)$
=$ 16 \hat j+ 8 \hat k$
$| \overrightarrow a \times \overrightarrow b|=\sqrt {(-16)^2+8^2}$
$\qquad\qquad=\sqrt {256+64}$
$\qquad\qquad=\sqrt {320}$
$\qquad\qquad=8\sqrt {5}$
$|\overrightarrow a|=\sqrt{3^2+1^2+2^2}=\sqrt{9+1+4}=\sqrt{14}$
$|\overrightarrow b|=\sqrt{(-2)^2+2^2+4^2}=\sqrt{4+4+16}=\sqrt{24}$
$\sin\theta = \large\frac{8 \sqrt 5}{4 \sqrt {21}}=\frac{2 \sqrt 5}{\sqrt {21}}$
$=\large\frac{2 \sqrt 5}{\sqrt {21}}$
answered May 27, 2013 by meena.p
 

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