A plane P is parallel to two lines whose direction ratios are $-2,1,-3$ and $-1,2,-2$ and it contains the point $(2,2,-2)$. Let P intersect the co-ordinate axes at the points $\mathrm{A}, \mathrm{B}, \mathrm{C}$ making the intercepts $\alpha, \beta, \gamma$. If $\mathrm{V}$ is the volume of the tetrahedron $\mathrm{OABC}$, where $\mathrm{O}$ is the origin, and $\mathrm{p}=\alpha+\beta+\gamma$, then the ordered pair $(\mathrm{V}, \mathrm{p})$ is equal to :
Solution
<p>Let ${\overrightarrow a _1} = ( - 2,1, - 3)$ and ${\overrightarrow a _2} = ( - 1,2, - 2)$</p>
<p>Vector normal to plane $\overline n = {\overrightarrow a _1} \times {\overrightarrow a _2}$</p>
<p>$\overline n = (4, - 1, - 3)$</p>
<p>Plane through $(2,2, - 2)$ and normal to $\overline n$</p>
<p>$(x - 2,y - 2,z + 2)\,.\,(4, - 1, - 3) = 0$</p>
<p>$\Rightarrow 4x - y - 3z = 12$</p>
<p>$\Rightarrow {x \over 3} + {y \over { - 12}} + {z \over { - 4}} = 1$</p>
<p>Intercepts $\alpha$, $\beta$, $\gamma$ are $3, - 12, - 4$</p>
<p>$P = \alpha + \beta + \gamma = - 13$</p>
<p>$V = {1 \over 6} \times 3 \times 12 \times 4 = 24$</p>
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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