If the equation of a plane P, passing through the intersection of the planes,
x + 4y - z + 7 = 0
and 3x + y + 5z = 8 is ax + by + 6z = 15 for some a, b $\in$ R, then the distance of the point
(3, 2, -1) from the plane P is...........
Answer (integer)
3
Solution
Equation of plane P is<br><br>$(x + 4y - z + 7) + \lambda (3x + y + 5z - 8) = 0$<br><br>$$ \Rightarrow x(1 + 3\lambda ) + y(4 + \lambda ) + z( - 1 + 5\lambda ) + (7 - 8\lambda ) = 0$$<br><br>$${{1 + 3\lambda } \over a} = {{4 + \lambda } \over b} = {{5\lambda - 1} \over 6} = {{7 - 8\lambda } \over { - 15}}$$
<br><br>$\therefore$ 15 - 75$\lambda$ = 42 - 48$\lambda$
<br><br>$\Rightarrow$ -27 = 27$\lambda$
<br><br>$\Rightarrow$ $\lambda$ = -1
<br><br>$\therefore$ Plane is $(x + 4y - z + 7) - 1 (3x + y + 5z - 8) = 0$
<br><br>$\Rightarrow$ $2x - 3y + 6z - 15 = 0$
<br><br>Distance of (3, 2, -1) from the plane P
<br><br>= ${{\left| {6 - 6 - 6 - 15} \right|} \over 7} = {{21} \over 7} = 3$
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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