The coordinates of the foot of the perpendicular from the point (1,-2,1) on the plane cont…

Question

The coordinates of the foot of the perpendicular from the point $(1,-2,1)$ on the plane containing the lines $\dfrac{x+1}6=\dfrac{y-1}7=\dfrac{z-3}8$ and $\dfrac{x-1}3=\dfrac{y-2}5=\dfrac{z-3}7$, is:

Accepted Answer

$$\overrightarrow n $$ = $$\overrightarrow {{n_1}}  \times \overrightarrow {{n_2}} $$ = $$\left| {\begin{pmatrix} {\widehat i} & {\widehat j} & {\widehat k}  \
   6 & 7 & 8  \
   3 & 5 & 7  \

\end{pmatrix} } \right|$$

= (9, - 18, 9)

= (1, -2, 1)

$$ \therefore $$   Equation of plane is

1(x + 1) - 2(y - 1) + (z - 3) = 0

$$ \Rightarrow $$   x - 2y + z = 0

foot to z

$${{x - 1} \over 1}$$ = $${{y + 2} \over { - 2}}$$ = $${{z - 1} \over 1}$$ = $$ - {{\left[ {1 + 4 + 1} \right]} \over 6}$$

x = 0,   y = 0,   z = 0

Solutions · 1