Let a line passing through the point P(4,1,0) | vedclass

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(A) Let the line $L_1$ be $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=p$. Then $A = (2p+1, 3p+2, 4p+3)$.Let the line $L_2$ be $\frac{x-6}{1}=\frac{y}{1

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$8$

Vedclass · Answer

(A) Let the line $L_1$ be $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=p$. Then $A = (2p+1, 3p+2, 4p+3)$.Let the line $L_2$ be $\frac{x-6}{1}=\frac{y}{1}=\frac{z-4}{-1}=q$. Then $B = (q+6, q, 4-q)$.The points $P(4,1,0)$,$A$,and $B$ are collinear. The direction ratios of $PA$ are $(2p+1-4, 3p+2-1, 4p+3-0) = (2p-3, 3p+1, 4p+3)$.The direction ratios of $AB$ are $(q+6-(2p+1), q-(3p+2), 4-q-(4p+3)) = (q-2p+5, q-3p-2, -q-4p+1)$.Since $P, A, B$ are collinear,the direction ratios of $PA$ and $AB$ are proportional:$\frac{2p-3}{q-2p+5} = \frac{3p+1}{q-3p-2} = \frac{4p+3}{-q-4p+1} = k$.Solving this system,we find $p=-1$ and $q=3$.Substituting $p=-1$ into $A$,we get $A(-1, -1, -1)$.Substituting $q=3$ into $B$,we get $B(9, 3, 1)$.Now,calculate the determinant:$\left|\begin{array}{ccc}1 & 0 & 1 \ -1 & -1 & -1 \ 9 & 3 & 1\end{array}ight| = 1(-1 - (-3)) - 0 + 1(-3 - (-9)) = 1(2) + 1(6) = 8$.

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