Let the equation of the plane passing through the line $x - 2y - z - 5 = 0 = x + y + 3z - 5$ and parallel to the line $x + y + 2z - 7 = 0 = 2x + 3y + z - 2$ be $ax + by + cz = 65$. Then the distance of the point (a, b, c) from the plane $2x + 2y - z + 16 = 0$ is ____________.
Answer (integer)
9
Solution
Let the equation of the plane is
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$(x-2 y-z-5)+\lambda(x+y+3 z-5)=0$
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$\because \quad$ it's parallel to the line
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$x+y+2 z-7=0=2 x+3 y+z-2$ <br/><br/>So, vector along the line $\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 2 \\ 2 & 3 & 1\end{array}\right|$
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$=-5 \hat{i}+3 \hat{j}+\hat{k}$
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$\because $ Plane is parallel to line
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$\therefore -5(1+\lambda)+3(-2+\lambda)+1(-1+3 \lambda)=0$
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$\lambda=12$
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So, by (i)
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$13 x+10 y+35 z=65$
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$\therefore a=13, b=10, c=35$
and $d=\frac{26+20-35+16}{\sqrt{9}}=9$
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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