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(C) The plane passes through the point $a$ and contains the line $r = b + \lambda c$. Therefore,the plane is parallel to the vector $c$ and the vector

Vedclass · Accepted Answer

$\frac{[a, b, c]}{|b 	imes c + c 	imes a|}$

Vedclass · Answer

(C) The plane passes through the point $a$ and contains the line $r = b + \lambda c$. Therefore,the plane is parallel to the vector $c$ and the vector $(b - a)$.The normal vector $n$ to the plane is given by the cross product of these two vectors:$n = (b - a) 	imes c = b 	imes c - a 	imes c = b 	imes c + c 	imes a$.The equation of the plane is $(r - a) \cdot n = 0$,which simplifies to $r \cdot n = a \cdot n$.Substituting $n = b 	imes c + c 	imes a$,we get $r \cdot (b 	imes c + c 	imes a) = a \cdot (b 	imes c + c 	imes a)$.Since $a \cdot (b 	imes c + c 	imes a) = a \cdot (b 	imes c) + a \cdot (c 	imes a) = [a, b, c] + 0 = [a, b, c]$,the equation of the plane is $r \cdot (b 	imes c + c 	imes a) = [a, b, c]$.The length of the perpendicular from the origin $(0, 0, 0)$ to the plane $r \cdot n = d$ is given by $\frac{|d|}{|n|}$.Here,$d = [a, b, c]$ and $n = b 	imes c + c 	imes a$.Thus,the length of the perpendicular is $\frac{[a, b, c]}{|b 	imes c + c 	imes a|}$.

The length of the perpendicular from the origin to the plane passing through the point $a$ and containing the line $r = b + \lambda c$ is

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