MHT CET · Maths · Vector Algebra
If \(\mathrm{P}(3,2,6), \mathrm{Q}(1,4,5)\) and \(\mathrm{R}(3,5,3)\) are the vertices of \(\Delta \mathrm{PQR}\), then \(\mathrm{m} \angle \mathrm{PQR}\) is
- A \(90^{\circ}\)
- B \(50^{\circ}\)
- C \(70^{\circ}\)
- D \(30^{\circ}\)
Answer & Solution
Correct Answer
(A) \(90^{\circ}\)
Step-by-step Solution
Detailed explanation
We have \(P \equiv(3,2,6) ; Q \equiv(1,4,5)\) and \(R \equiv(3,5,3)\)
d.r. of \(\mathrm{PQ}\) are \(-2,2,-1\) and d.r. of \(\mathrm{QR}\) are \(2,1,-2\)
We find that : \((-2)(2)+2(1)+(-1)(-2)=-4+2+2=0\)
\(\therefore \mathrm{PQ} \perp \mathrm{QR} \Rightarrow \mathrm{m} \angle \mathrm{PQR}=90^{\circ}\)
d.r. of \(\mathrm{PQ}\) are \(-2,2,-1\) and d.r. of \(\mathrm{QR}\) are \(2,1,-2\)
We find that : \((-2)(2)+2(1)+(-1)(-2)=-4+2+2=0\)
\(\therefore \mathrm{PQ} \perp \mathrm{QR} \Rightarrow \mathrm{m} \angle \mathrm{PQR}=90^{\circ}\)
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