The coordinates of the point where the line j | vedclass

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Question

(A) The equation of the $yz$-plane is $x = 0$. Let the point $P$ divide the line segment joining $A(3, 5, -7)$ and $B(-2, 1, 8)$ in the ratio $k:1$. T

Vedclass · Accepted Answer

$\left(0, \frac{13}{5}, 2\right)$

Vedclass · Answer

(A) The equation of the $yz$-plane is $x = 0$. Let the point $P$ divide the line segment joining $A(3, 5, -7)$ and $B(-2, 1, 8)$ in the ratio $k:1$. The coordinates of $P$ are given by $\left(\frac{-2k + 3}{k+1}, \frac{k + 5}{k+1}, \frac{8k - 7}{k+1}\right)$. Since $P$ lies on the $yz$-plane,its $x$-coordinate must be $0$: $\frac{-2k + 3}{k+1} = 0 \implies -2k + 3 = 0 \implies k = \frac{3}{2}$. Substituting $k = \frac{3}{2}$ into the coordinates of $P$: $y = \frac{\frac{3}{2} + 5}{\frac{3}{2} + 1} = \frac{\frac{13}{2}}{\frac{5}{2}} = \frac{13}{5}$. $z = \frac{8(\frac{3}{2}) - 7}{\frac{3}{2} + 1} = \frac{12 - 7}{\frac{5}{2}} = \frac{5}{\frac{5}{2}} = 2$. Thus,the coordinates are $\left(0, \frac{13}{5}, 2\right)$.

The coordinates of the point where the line joining the points $(3, 5, -7)$ and $(-2, 1, 8)$ is intersected by the $yz$-plane are given by:

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