If vectors \( \vec{a} = 2\hat{i} + 3\hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} - \hat{j} + 2\hat{k} \), then the magnitude of the projection of \( \vec{a} \) on \( \vec{b} \) is:
Show Hint
- \( \sqrt{3} \)
- \( \sqrt{6} \)
- \( 2 \)
- \( \sqrt{2} \)
The Correct Option is C
Solution and Explanation
Step 1: Use Scalar Projection Formula
The magnitude of the projection of \( \vec{a} \) on \( \vec{b} \) is: \[ \text{Projection} = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{b}|} \]
Step 2: Compute Dot Product
Let \( \vec{a} = 2\hat{i} + 3\hat{j} + \hat{k},\ \vec{b} = \hat{i} \). Then: \[ \vec{a} \cdot \vec{b} = 2(1) + 3(0) + 1(0) = 2 \]
Step 3: Compute Magnitude of \( \vec{b} \)
\[ |\vec{b}| = \sqrt{1^2 + 0^2 + 0^2} = 1 \]
Step 4: Final Answer
\[ \text{Scalar projection} = \frac{2}{1} = 2 \]
Top Questions on Vectors
- Let $|\vec{a}| = 5 \text{ and } -2 \leq \lambda \leq 1$. Then, the range of $|\lambda \vec{a}|$ is:
- Consider the vectors $$ \vec{x} = \hat{i} + 2\hat{j} + 3\hat{k},\quad \vec{y} = 2\hat{i} + 3\hat{j} + \hat{k},\quad \vec{z} = 3\hat{i} + \hat{j} + 2\hat{k}. $$ For two distinct positive real numbers $ \alpha $ and $ \beta $, define $$ \vec{X} = \alpha \vec{x} + \beta \vec{y} - \vec{z},\quad \vec{Y} = \alpha \vec{y} + \beta \vec{z} - \vec{x},\quad \vec{Z} = \alpha \vec{z} + \beta \vec{x} - \vec{y}. $$ If the vectors $ \vec{X}, \vec{Y}, \vec{Z} $ lie in a plane, then the value of $ \alpha + \beta - 3 $ is ________.
Let $ \vec{w} = \hat{i} + \hat{j} - 2\hat{k} $, and $ \vec{u} $ and $ \vec{v} $ be two vectors, such that $ \vec{u} \times \vec{v} = \vec{w} $ and $ \vec{v} \times \vec{w} = \vec{u} $. Let $ \alpha, \beta, \gamma $, and $ t $ be real numbers such that: $$ \vec{u} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}, $$ and the system of equations is: $$ -t\alpha + \beta + \gamma = 0 \quad \cdots (1) $$ $$ \alpha - t\beta + \gamma = 0 \quad \cdots (2) $$ $$ \alpha + \beta - t\gamma = 0 \quad \cdots (3) $$ Match each entry in List-I to the correct entry in List-II and choose the correct option.
List-I- [(P)] $ |\vec{v}|^2 $ is equal to
- [(Q)] If $ \alpha = \sqrt{3} $, then $ \gamma^2 $ is equal to
- [(R)] If $ \alpha = \sqrt{3} $, then $ (\beta + \gamma)^2 $ is equal to
- [(S)] If $ \alpha = \sqrt{2} $, then $ t + 3 $ is equal to
List-II
- [(1)] 0
- [(2)] 1
- [(3)] 2
- [(4)] 3
- [(5)]
- The diagonals of a parallelogram are given by \( \mathbf{a} = 2 \hat{i} - \hat{j} + \hat{k} \) and \( \mathbf{b} = \hat{i} + 3 \hat{j} - \hat{k}\) . Find the area of the parallelogram.
- Let \( L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} \) and \( L_2: \frac{x-2}{3} = \frac{y-4}{4} = \frac{z-5}{5} \) be two lines. Then which of the following points lies on the line of the shortest distance between \( L_1 \) and \( L_2 \)?
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