JAMIA 2021 — Mathematics PYQ

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Question

JAMIA 2021 — Mathematics PYQ

JAMIA | Mathematics | 2021

For any vector $a$ , the value of $(a \times \hat{i})^{2} + a \times \hat{j})^{2} + (a \times$ $\hat{k})^{2}$ is equal to

Choose the correct answer:

Explanation

1. Let the vector $\vec{a}$ be:

\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}

2. Find the square of the magnitude $|\vec{a}|^{2}$ :

|\vec{a}|^{2} = a^{2} = x^{2} + y^{2} + z^{2}

3. Calculate the cross product $(\vec{a} \times \hat{i})$ :

\vec{a} \times \hat{i} = (x\hat{i} + y\hat{j} + z\hat{k}) \times \hat{i}

\vec{a} \times \hat{i} = x(\hat{i} \times \hat{i}) + y(\hat{j} \times \hat{i}) + z(\hat{k} \times \hat{i})

\vec{a} \times \hat{i} = 0 - y\hat{k} + z\hat{j} = z\hat{j} - y\hat{k}

4. Find the squared magnitude $(\vec{a} \times \hat{i})^{2}$ :

(\vec{a} \times \hat{i})^{2} = |z\hat{j} - y\hat{k}|^{2} = z^{2} + y^{2}

5. Similarly, find the values for $\hat{j}$ and $\hat{k}$ :

(\vec{a} \times \hat{j})^{2} = |(x\hat{i} + y\hat{j} + z\hat{k}) \times \hat{j}|^{2} = |x\hat{k} - z\hat{i}|^{2} = x^{2} + z^{2}

(\vec{a} \times \hat{k})^{2} = |(x\hat{i} + y\hat{j} + z\hat{k}) \times \hat{k}|^{2} = |-x\hat{j} + y\hat{i}|^{2} = x^{2} + y^{2}

6. Add the three squared terms together:

(\vec{a} \times \hat{i})^{2} + (\vec{a} \times \hat{j})^{2} + (\vec{a} \times \hat{k})^{2} = (z^{2} + y^{2}) + (x^{2} + z^{2}) + (x^{2} + y^{2})

= 2x^{2} + 2y^{2} + 2z^{2}

= 2(x^{2} + y^{2} + z^{2})

Final Answer:

Since $x^{2} + y^{2} + z^{2} = |\vec{a}|^{2}$ , the value is:

2|\vec{a}|^{2}