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The vectors $\overrightarrow{a}=3\hat i+2\hat j+2\hat k$ and $\overrightarrow{b}=-\hat i+2\hat k$ are the adjacent sides of a parallelogram .The acute angle between its diagonals is__________.

$\begin{array}{1 1}(A)\;\sin ^{-1} \bigg(\sqrt {\large\frac{84}{85}}\bigg) \\(B)\; \cos^{-1}\large\frac{1}{\sqrt{85}} \\(C)\;\cos ^{-1} \bigg(\sqrt {\large\frac{85}{84}}\bigg) \\(D)\; \sin^{-1}\large\frac{4}{\sqrt{30}} \end{array} $

1 Answer

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  • $\sin \theta=\large\frac{|\overrightarrow a \times \overrightarrow b|}{|\overrightarrow a||\overrightarrow b|}$
Let $\overrightarrow a=3 \hat i+2 \hat j+2 \hat k$ and $\overrightarrow b=-\hat i+2 \hat k$
Given $\overrightarrow a$ and $\overrightarrow b$ are the adjacent sides of the parallelogram.
Therefore $\sin \theta=\large\frac{|\overrightarrow a \times \overrightarrow b|}{|\overrightarrow a||\overrightarrow b|}$
$|\overrightarrow a|=\sqrt {(3)^2+(2)^2+(2)^2}$
$\qquad= \sqrt {9+4+4}$
$\qquad= \sqrt {17}$
$|\overrightarrow b|=\sqrt {(-1)^2+(2)^2}$
$\qquad= \sqrt {1+4}$
$\qquad= \sqrt {5}$
$\overrightarrow a \times \overrightarrow b$$=\begin {vmatrix} \hat i & \hat j & \hat k \\ 3 & 2 & 2 \\ -1 & 0 & 2 \end {vmatrix}$
$\qquad\quad=\hat i(4-0)-\hat j(6+2)+\hat k(0+2)$
$\qquad\quad=4 \hat i-8 \hat j+2 \hat k$
$|\overrightarrow a \times \overrightarrow b| =\sqrt {4^2+8^2+2^2}$
$\qquad\qquad=\sqrt {84}$
Now substituting the values
Therefore $\sin \theta=\large\frac{\sqrt {84}}{\sqrt {17} \sqrt 5}$
$\qquad\qquad\qquad=\large\frac{\sqrt {84}}{\sqrt {85}}$
Therefore $\theta=\sin ^{-1} \bigg(\sqrt {\large\frac{84}{85}}\bigg)$
answered May 30, 2013 by meena.p
 

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