Question:medium

For vectors \( \vec{a}=3\hat{i}-\hat{j}+2\hat{k} \) and \( \vec{b}=\hat{i}+2\hat{j}-\hat{k} \), find the scalar projection of \( \vec{a} \) onto \( \vec{b} \).

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A negative scalar projection means the vector has a component opposite to the direction of the reference vector.
Updated On: Jun 3, 2026
  • \( -\frac1{\sqrt6} \)
  • \( \frac1{\sqrt6} \)
  • \( -\frac16 \)
  • \( \frac16 \)
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
The scalar projection of a vector $\vec{a}$ onto another vector $\vec{b}$ represents the directional shadow length or component magnitude of vector $\vec{a}$ along the line of action of vector $\vec{b}$. Geometrically, it equals the magnitude of $\vec{a}$ multiplied by the cosine of the angle $\theta$ between the two vectors ($|\vec{a}|\cos\theta$). A negative scalar projection value simply means that the vector component points in the exact opposite direction of vector $\vec{b}$.
Step 2: Key Formula or Approach:
The mathematical formula used to calculate the scalar projection of vector $\vec{a}$ onto vector $\vec{b}$ is: $$ \text{Scalar Projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} $$ Where: - $\vec{a} \cdot \vec{b}$ is the standard vector dot product (scalar product). - $|\vec{b}|$ is the geometric magnitude length of the target baseline vector $\vec{b}$.
Step 3: Detailed Explanation:
Let's calculate the numerical components step-by-step using our given vector values: $$ \vec{a} = 3\hat{i} - \hat{j} + 2\hat{k} $$ $$ \vec{b} = \hat{i} + 2\hat{j} - \hat{k} $$ 1. Step 1: Compute the dot product ($\vec{a} \cdot \vec{b}$): Multiply the corresponding directional components ($\hat{i}, \hat{j}, \hat{k}$) of both vectors together and sum the results: $$ \vec{a} \cdot \vec{b} = (3 \times 1) + (-1 \times 2) + (2 \times -1) $$ $$ \vec{a} \cdot \vec{b} = 3 - 2 - 2 = -1 $$ 2. Step 2: Compute the magnitude of the target base vector ($|\vec{b}|$): Take the square root of the sum of the squares of its components: $$ |\vec{b}| = \sqrt{(1)^2 + (2)^2 + (-1)^2} $$ $$ |\vec{b}| = \sqrt{1 + 4 + 1} = \sqrt{6} $$ 3. Step 3: Combine the components into our projection template: $$ \text{Scalar Projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{-1}{\sqrt{6}} $$ This calculated value matches option (A).
Step 4: Final Answer:
The scalar projection of $\vec{a}$ onto $\vec{b}$ is -1/$\sqrt{6}$.
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