JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let a line passing through the point \((4,1,0)\) intersect the line \(L_1 ; \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) at the point \(\mathrm{A} \quad(\alpha, \beta, \gamma)\) and the line \(L_2: x-6=y=-z+4\) at the point \(B(a, b, c)\).
Then \(\left|\begin{array}{lll}1 & 0 & 1 \\ \alpha & \beta & \gamma \\ \mathrm{a} & \mathrm{b} & \mathrm{c}\end{array}\right|\) is equal to
- A \(8\)
- B \(16\)
- C \(12\)
- D \(6\)
Answer & Solution
Correct Answer
(A) \(8\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & L_1=\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=p \\ & L_2=\frac{x-6}{1}=\frac{y}{1}=\frac{z-4}{-1}=q \\ & A(2 p+1,3 p+2,4 p+3) \\ & B(q+6, q, 4-q) \\ & \text { D.R. of } P A=2 p-3,3 P+1,4 p+3 \\ & \text { D.R. of } P B=q+2, q-1,4-q \\ & \frac{2 p-3}{q+2}=\frac{3…
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