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$L$ is line of intersection of the planes $2x+3y+z=1\:\:and\:\:x+3y+2z=2$ and $\alpha$ is the angle made by the line $L$ with positive $x-axis$ then $cos\alpha=?$

$\begin{array}{1 1} (a)\:\:\large\frac{1}{2}\:\:\:\qquad\:\:(b)\:\:1\:\:\:\qquad\:\:(c)\:\:\frac{1}{\sqrt 2}\:\:\qquad\:\:(d)\frac{1}{\sqrt 3}. \end{array} $</p

1 Answer

  • Direction cosine, $d.c$ of any vector $x\hat i+y\hat j+z\hat k$ is $\bigg(\large\frac{x}{\sqrt {x^2+y^2+z^2}},\frac{y}{\sqrt {x^2+y^2+z^2}},\frac{z}{\sqrt {x^2+y^2+z^2}}\bigg)$
Let the direction ratio $(d.r.)$ of the line of intersection $L$ of the planes
$2x+3y+z-1=0..........(i)\:\:and\:\:x+3y+2z-2=0......(ii)$ be $(l,m,n)$
Since $L$ lies on both the planes, it is $|perp$ to the normals to both the planes.
We know that any vector $\perp$ to $\overrightarrow {n_1}\:\:and\:\:\overrightarrow {n_2}$ is $\overrightarrow {n_1}\times\overrightarrow {n_2}$.
Normal to the plane $(i)\:\:\overrightarrow {n_1}=(2,3,1)$ and that of plane $(ii)\:\:\overrightarrow {n_2}\:\:is\:\:(1,3,2)$
$\overrightarrow {n_1}\times\overrightarrow {n_2}=\left |\begin {array}{ccc}\hat i & \hat j & \hat k \\ 2 & 3 & 1 \\1 & 3 & 2\end {array}\right | = (3,-3,3)$
$\therefore\:d.c.$ of the line $L$ = $\big(\large\frac{1}{\sqrt 3},-\frac{1}{\sqrt 3},\frac{1}{\sqrt 3}\big)$
$\Rightarrow\:$ $cos\alpha = \large\frac{1}{\sqrt 3}$.
answered Dec 28, 2013 by rvidyagovindarajan_1

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