The orthogonal projection vector of \( \bar{a} = 2\bar{i} + 3\bar{j} + 3\bar{k} \) on \( \bar{b} = \bar{i} - 2\bar{j} + \bar{k} \) is:
Show Hint
- \( -\frac{1}{6} (2\bar{i} + 3\bar{j} + 3\bar{k}) \)
- \( \frac{1}{6} (-\bar{i} + 2\bar{j} - \bar{k}) \)
- \( \bar{i} - 2\bar{j} + \bar{k} \)
- \( -\bar{i} + 2\bar{j} - \bar{k} \)
The Correct Option is B
Solution and Explanation
Step 1: Formula for vector projection \[ \text{Proj}_{\bar{b}} \bar{a} = \frac{\bar{a} \cdot \bar{b}}{\bar{b} \cdot \bar{b}} \bar{b}. \] Step 2: Compute dot products \[ \bar{a} \cdot \bar{b} = (2)(1) + (3)(-2) + (3)(1) = 2 - 6 + 3 = -1. \] \[ \bar{b} \cdot \bar{b} = (1)^2 + (-2)^2 + (1)^2 = 1 + 4 + 1 = 6. \] Step 3: Compute projection \[ \text{Proj}_{\bar{b}} \bar{a} = \frac{-1}{6} (1\bar{i} - 2\bar{j} + 1\bar{k}) = \frac{1}{6} (-\bar{i} + 2\bar{j} - \bar{k}). \]
Top Questions on Vectors
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- Let \(\vec{a} = i + j + k\), \(\vec{b} = 2i + 2j + k\) and \(\vec{d} = \vec{a} \times \vec{b}\). If \(\vec{c}\) is a vector such that \(\vec{a} \cdot \vec{c} = |\vec{c}|\), \(\|\vec{c} - 2\vec{d}\| = 8\) and the angle between \(\vec{d}\) and \(\vec{c}\) is \(\frac{\pi}{4}\), then \(\left|10 - 3\vec{b} \cdot \vec{c} + |\vec{d}|\right|^2\) is equal to:
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