Vector Algebra
Dot Product and Projection
Grade 12

Question:

<p>The projection of the vector $\vec{i} - \vec{j}$ on the vector $\vec{i} + \vec{j}$ is</p>
<p>(a) 0</p>
<p>(b) 1</p>
<p>(c) $\frac{1}{2}$</p>
<p>(d) None of these</p>

Step-by-Step Solution

Key Concept: Use the dot product formula for projection: projection of vector a on b equals (a·b)/|b|.
Solution: Let $\vec{a} = \hat{i} - \hat{j}$ and $\vec{b} = \hat{i} + \hat{j}$ Projection of $\vec{a}$ on $\vec{b}$ is given by: $\text{Projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{(\hat{i} - \hat{j}) \cdot (\hat{i} + \hat{j})}{\sqrt{1^2 + 1^2}} = \frac{1 \cdot 1 + (-1) \cdot 1}{\sqrt{2}} = \frac{1 - 1}{\sqrt{2}} = 0$ Hence, the projection of vector $\vec{a}$ on $\vec{b}$ is 0.
Correct Answer: a

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