Question:
Let O be the origin and the position vector of the point P be \(\hat i-2\hat j+3\hat k\). If the position vectors of the points A, B and C are \(-2\hat i+\hat j-3\hat k,2\hat i+4\hat j-2\hat k\) and \(-4\hat i+2\hat j-\hat k\) respectively then the projection of the vector \(\overrightarrow{OP}\) on a vector perpendicular to the vectors \(\overrightarrow{ AB}\) and \(\overrightarrow {AC}\) is
Let O be the origin and the position vector of the point P be \(\hat i-2\hat j+3\hat k\). If the position vectors of the points A, B and C are \(-2\hat i+\hat j-3\hat k,2\hat i+4\hat j-2\hat k\) and \(-4\hat i+2\hat j-\hat k\) respectively then the projection of the vector \(\overrightarrow{OP}\) on a vector perpendicular to the vectors \(\overrightarrow{ AB}\) and \(\overrightarrow {AC}\) is
Updated On: Mar 11, 2026
- \(\frac{7}{3}\)
- \(\frac{8}{3}\)
- 3
- \(\frac{10}{3}\)
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The Correct Option is C
Solution and Explanation
Given Points: \( P(-1, -2, 3), \, A(-2, 1, -3), \, B(2, 4, -2), \, C(-4, 2, -1) \)
To Find: Position vector \( \overrightarrow{OP} \)
- Step 1: Find \( \overrightarrow{AB} \times \overrightarrow{AC} \):
- \( \overrightarrow{AB} = \langle 4, 3, 1 \rangle, \quad \overrightarrow{AC} = \langle -2, 1, 2 \rangle \)
- \( \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 3 & 1 \\ -2 & 1 & 2 \end{vmatrix} \)
- Expanding the determinant:
\( \overrightarrow{AB} \times \overrightarrow{AC} = \hat{i}(3 \cdot 2 - 1 \cdot 1) - \hat{j}(4 \cdot 2 - 1 \cdot -2) + \hat{k}(4 \cdot 1 - 3 \cdot -2) \)
- \( \overrightarrow{AB} \times \overrightarrow{AC} = \hat{i}(5) - \hat{j}(8 + 2) + \hat{k}(4 + 6) \)
- \( \overrightarrow{AB} \times \overrightarrow{AC} = 5\hat{i} - 10\hat{j} + 10\hat{k} \)
- Step 2: Compute \( \overrightarrow{OP} \):
- \( \overrightarrow{OP} = \frac{\overrightarrow{AB} \times \overrightarrow{AC}}{\lvert \overrightarrow{AB} \times \overrightarrow{AC} \rvert} \)
- Magnitude of \( \overrightarrow{AB} \times \overrightarrow{AC} \):
\( \lvert \overrightarrow{AB} \times \overrightarrow{AC} \rvert = \sqrt{5^2 + (-10)^2 + 10^2} \)
- \( \lvert \overrightarrow{AB} \times \overrightarrow{AC} \rvert = \sqrt{25 + 100 + 100} = \sqrt{225} = 15 \)
- Direction Cosine:
- \( \overrightarrow{OP} = \frac{5\hat{i} - 10\hat{j} + 10\hat{k}}{15} \)
- \( \overrightarrow{OP} = \hat{i}(-1) - \hat{j}(2) + \hat{k}(3) \)
- Simplify:
\( \overrightarrow{OP} = \frac{-5 + 20 + 30}{\sqrt{25 + 100 + 100}} = \frac{45}{15} = 3 \)
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