TS EAMCET · Maths · Vector Algebra
The position vectors of two points A and B are \(\bar{i}+2 \bar{j}+3 \bar{k}\) and \(7 \bar{i}-\bar{k}\) respectively. The point P with position vector \(-2 \bar{i}+3 \bar{j}+5 \bar{k}\) is on the line AB. If the point Q is the harmonic conjugate of P, then the sum of the scalar components of the position vector of Q is
- A \(6\)
- B \(4\)
- C \(2\)
- D \(0\)
Answer & Solution
Correct Answer
(A) \(6\)
Step-by-step Solution
Detailed explanation
\( \vec{A} = \bar{i}+2 \bar{j}+3 \bar{k}, \vec{B} = 7 \bar{i}-\bar{k}, \vec{P} = -2 \bar{i}+3 \bar{j}+5 \bar{k} \) P divides AB in ratio \( \lambda:1 \): \( \vec{P} = \frac{\lambda \vec{B} + \vec{A}}{\lambda+1} \) Equating \( \bar{i} \) components:…
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