Class CBSE Class 12 Mathematics Vector Algebra Q #1377
KNOWLEDGE BASED
REMEMBER
2 Marks 2025 AISSCE(Board Exam) VSA
A vector $\vec{a}$ makes equal angles with all the three axes. If the magnitude of the vector is $5\sqrt{3}$ units, then find $\vec{a}$.
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Detailed Solution

Step 1: Define direction cosines

Let $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$. Since $\vec{a}$ makes equal angles with all three axes, its direction cosines are equal. Let the angle be $\alpha$. Then, $l = \cos\alpha$, $m = \cos\alpha$, and $n = \cos\alpha$.

Step 2: Use the property of direction cosines

We know that $l^2 + m^2 + n^2 = 1$. Substituting the values, we get: $(\cos\alpha)^2 + (\cos\alpha)^2 + (\cos\alpha)^2 = 1$ $3\cos^2\alpha = 1$ $\cos^2\alpha = \frac{1}{3}$ $\cos\alpha = \pm\frac{1}{\sqrt{3}}$

Step 3: Find the direction ratios

The direction ratios are proportional to the direction cosines. Therefore, $a_1 = a_2 = a_3 = k$, where $k$ is a constant. So, $\vec{a} = k\hat{i} + k\hat{j} + k\hat{k}$.

Step 4: Use the magnitude of the vector

Given that $|\vec{a}| = 5\sqrt{3}$. We have: $\sqrt{k^2 + k^2 + k^2} = 5\sqrt{3}$ $\sqrt{3k^2} = 5\sqrt{3}$ $|k|\sqrt{3} = 5\sqrt{3}$ $|k| = 5$ $k = \pm 5$

Step 5: Determine the vector a

Therefore, $\vec{a} = \pm 5(\hat{i} + \hat{j} + \hat{k})$ $\vec{a} = 5\hat{i} + 5\hat{j} + 5\hat{k}$ or $\vec{a} = -5\hat{i} - 5\hat{j} - 5\hat{k}$

Final Answer: $\vec{a} = \pm 5(\hat{i} + \hat{j} + \hat{k})$

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Pedagogical Audit
Bloom's Analysis: This is an REMEMBER question because it requires recalling the formula for direction cosines and the relationship between the magnitude of a vector and its components.
Knowledge Dimension: CONCEPTUAL
Justification: The question requires understanding the concepts of direction cosines, direction ratios, and vector magnitude. It involves applying these concepts to find the vector.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. It directly tests the student's understanding of vectors and their properties as covered in the textbook.

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