find the point of intersection of the line r^→ =(1−2t,3+4t,t) and the plane 3x−2y+5z=15


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Question Number 110320 by bemath last updated on 28/Aug/20

$f i n d t h e p o i n t o f i n t e r s e c t i o n$ $o f t h e l i n e \vec{r} = (1 - 2 t, 3 + 4 t, t)$ $a n d t h e p l a n e 3 x - 2 y + 5 z = 15$
	Answered by mr W last updated on 28/Aug/20

	$3 (1 - 2 t) - 2 (3 + 4 t) + 5 (t) = 15$ $- 9 t = 18$ $\Rightarrow t = - 2$ $1 - 2 t = 5$ $3 + 4 t = - 5$ $i n t e r s e c t i o n p o i n t i s (5, - 5, - 2)$
	Commented by I want to learn more last updated on 28/Aug/20

	$Sir please help me check the permutation question in Q 110156 and 110157$
	Answered by Rio Michael last updated on 28/Aug/20

	$x = 1 - 2 t$ $y = 3 + 4 t$ $z = t$ $substituting \Rightarrow 3 (1 - 2 t) - 2 (3 + 4 t) + 5 t = 15$ $3 - 6 t - 6 - 8 t + 5 t = 15$ $\Rightarrow - 9 t = 18 hence t = - 2$ $\Rightarrow point of intersection is (5, - 5, - 2)$
	Answered by 1549442205PVT last updated on 28/Aug/20

	$The equation of the line is {\begin{cases} x = 1 - 2 t \\ y = 3 + 4 t \\ z = t \end{cases}$ $Replace into 3 x - 2 y + 5 z = 15 we have$ $3 (1 - 2 t) - 2 (3 + 4 t) + 5 t = 15$ $\Leftrightarrow - 9 t = 18 \Rightarrow t = - 2$ $Substituting into the system above$ $we get x = 5, y = - 5, - 2$ $Therefore, the intersection point of$ $the line and the plane is the point$ $M (5, - 5, - 2)$
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