Question

If \(\vec{a}\) is any vector, then the value of \(|\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2\) is equal to

• A. \(2|\vec{a}|^2\)
• B. \(|\vec{a}|^2\)
• C. \(4|\vec{a}|^2\)
• D. \(3|\vec{a}|^2\)

Hint:

Use \(|\vec{a}\times\vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a}\cdot\vec{b})^2\) for each unit vector.

Answer 1

The Correct Option is A

Step 1: Use the Lagrange identity.
For any two vectors, \(|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2\).

Step 2: Apply it to each unit vector.
\(|\vec{a} \times \hat{i}|^2 = |\vec{a}|^2 - a_1^2\), \(|\vec{a} \times \hat{j}|^2 = |\vec{a}|^2 - a_2^2\), \(|\vec{a} \times \hat{k}|^2 = |\vec{a}|^2 - a_3^2\). Here \(\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}\) and each unit vector has length 1.

Step 3: Add.
\[ 3|\vec{a}|^2 - (a_1^2 + a_2^2 + a_3^2) = 3|\vec{a}|^2 - |\vec{a}|^2 = 2|\vec{a}|^2 \]

Step 4: Check the options.
Option 2 would need the sum to be one copy of \(|\vec{a}|^2\), and option 4 forgets to subtract the dot product terms. Option 3 is too large. Option 1 is right.

Final Answer:
The value is \(2|\vec{a}|^2\), option 1. \[ \boxed{2|\vec{a}|^2} \]