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(B) Let the $yz$-plane divide the line segment joining the points $(a, 1, 6)$ and $(3, 4, b)$ in the ratio $\lambda : 1$. Using the section formula,th

Vedclass · Accepted Answer

$a=5, b=1$

Vedclass · Answer

(B) Let the $yz$-plane divide the line segment joining the points $(a, 1, 6)$ and $(3, 4, b)$ in the ratio $\lambda : 1$. Using the section formula,the point of intersection is given by: $\left(\frac{3\lambda + a}{\lambda + 1}, \frac{4\lambda + 1}{\lambda + 1}, \frac{b\lambda + 6}{\lambda + 1}ight) = \left(0, \frac{17}{2}, \frac{-13}{2}ight)$. Equating the coordinates: $1) \frac{3\lambda + a}{\lambda + 1} = 0 \Rightarrow 3\lambda + a = 0 \Rightarrow a = -3\lambda$. $2) \frac{4\lambda + 1}{\lambda + 1} = \frac{17}{2} \Rightarrow 8\lambda + 2 = 17\lambda + 17 \Rightarrow -9\lambda = 15 \Rightarrow \lambda = -\frac{15}{9} = -\frac{5}{3}$. Substituting $\lambda = -\frac{5}{3}$ into $a = -3\lambda$: $a = -3\left(-\frac{5}{3}ight) = 5$. $3) \frac{b\lambda + 6}{\lambda + 1} = -\frac{13}{2} \Rightarrow \frac{b(-\frac{5}{3}) + 6}{-\frac{5}{3} + 1} = -\frac{13}{2} \Rightarrow \frac{-\frac{5b}{3} + 6}{-\frac{2}{3}} = -\frac{13}{2}$. Multiply both sides by $-\frac{2}{3}$: $-\frac{5b}{3} + 6 = \left(-\frac{13}{2}ight) 	imes \left(-\frac{2}{3}ight) = \frac{13}{3}$. $-\frac{5b}{3} = \frac{13}{3} - 6 = \frac{13 - 18}{3} = -\frac{5}{3}$. $b = 1$. Thus,$a = 5$ and $b = 1$.

If the line passing through the points $(a, 1, 6)$ and $(3, 4, b)$ crosses the $yz$-plane at the point $\left(0, \frac{17}{2}, \frac{-13}{2}\right)$,then:

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