JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the point, on the line passing through the points \(P(1,-2,3)\) and \(Q(5,-4,7)\), farther from the origin and at a distance of \(9\) units from the point \(\mathrm{P}\), be \((\alpha, \beta, \gamma)\). Then \(\alpha^2+\beta^2+\gamma^2\) is equal to :
- A \(155\)
- B \(150\)
- C \(160\)
- D \(165\)
Answer & Solution
Correct Answer
(A) \(155\)
Step-by-step Solution
Detailed explanation
\(PQ\) line \( \frac{\mathrm{x}-1}{4}=\frac{\mathrm{y}+2}{-2}=\frac{\mathrm{z}-3}{4} \) \( \mathrm{pt}(4 \mathrm{t}+1,-2 \mathrm{t}-2,4 \mathrm{t}+3) \) \( \text { distance }^2=16 \mathrm{t}^2+4 \mathrm{t}^2+16 \mathrm{t}^2=81 \) \( \mathrm{t}= \pm \frac{3}{2} \)…
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