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Equation of a Plane : Vector Form

Consider a pl...

Question

Consider a plane $x + y - z = 1$ and point $A (1, 2, - 3) .$ A line $L$ has the equation $x = 1 + 3 r, y = 2 - r$ and $z = 3 + 4 r .$

The equation of the plane containing line $L$ and point $A$ has the equation

x - 3 y + 5 = 0

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x + 3 y - 7 = 0

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3 x - y - 1 = 0

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3 x + y - 5 = 0

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Solution

The correct option is B $x + 3 y - 7 = 0$
The equation of plane containing the line $L$ is,
$A (x - 1) + B (y - 2) + C (z - 3) = 0,$
where $3 A - B + 4 C = 0 . . . (1)$

Eqn $(1)$ also contains point $A (1, 2, - 3) .$
Hence, $C = 0$ and $3 A = B$

The equation of plane is,
$x - 1 + 3 (y - 2) = 0$
$\Rightarrow x + 3 y - 7 = 0$

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Consider a plane x+y−z=1 and point A(1,2,−3). A line L has the equation x=1+3r,y=2−r and z=3+4r. The equation of the plane containing line L and point A has the equation

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Consider a plane $x + y - z = 1$ and point $A (1, 2, - 3) .$ A line $L$ has the equation $x = 1 + 3 r, y = 2 - r$ and $z = 3 + 4 r .$

The equation of the plane containing line $L$ and point $A$ has the equation