P, Q, R and S are four points with the positi | vedclass

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(B) Let the coordinates of the four points $P, Q, R$ and $S$ be $(3, -4, 5), (0, 0, 4), (-4, 5, 1)$ and $(-3, 4, 3)$ respectively.The equation of line

Vedclass · Accepted Answer

$-3i+4j+3k$

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(B) Let the coordinates of the four points $P, Q, R$ and $S$ be $(3, -4, 5), (0, 0, 4), (-4, 5, 1)$ and $(-3, 4, 3)$ respectively.The equation of line $PQ$ passing through $(3, -4, 5)$ and $(0, 0, 4)$ is given by:$\frac{x-3}{0-3} = \frac{y+4}{0+4} = \frac{z-5}{4-5} = r_1$$\Rightarrow \frac{x-3}{-3} = \frac{y+4}{4} = \frac{z-5}{-1} = r_1$The equation of line $RS$ passing through $(-4, 5, 1)$ and $(-3, 4, 3)$ is given by:$\frac{x+4}{-3+4} = \frac{y-5}{4-5} = \frac{z-1}{3-1} = r_2$$\Rightarrow \frac{x+4}{1} = \frac{y-5}{-1} = \frac{z-1}{2} = r_2$Let a general point on line $PQ$ be $(-3r_1+3, 4r_1-4, -r_1+5)$ and on line $RS$ be $(r_2-4, -r_2+5, 2r_2+1)$.Since the lines intersect,we equate the coordinates:$-3r_1+3 = r_2-4 \Rightarrow 3r_1+r_2 = 7$ (Eq. $i$)$4r_1-4 = -r_2+5 \Rightarrow 4r_1+r_2 = 9$ (Eq. $ii$)Subtracting Eq. $i$ from Eq. $ii$:$(4r_1+r_2) - (3r_1+r_2) = 9 - 7 \Rightarrow r_1 = 2$Substituting $r_1 = 2$ in Eq. $i$:$3(2) + r_2 = 7 \Rightarrow r_2 = 1$Substituting $r_2 = 1$ into the coordinates of line $RS$:$x = 1-4 = -3, y = -1+5 = 4, z = 2(1)+1 = 3$Thus,the point of intersection is $(-3, 4, 3)$,which corresponds to $-3i+4j+3k$.

$P, Q, R$ and $S$ are four points with the position vectors $3i-4j+5k, 0i+0j+4k, -4i+5j+1k$ and $-3i+4j+3k$,respectively. Then,the line $PQ$ meets the line $RS$ at the point

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