Question

If \( \theta \) is the angle between two vectors \( \vec{a} \) and \( \vec{b} \) such that \( |\vec{a}| = 7 \), \( |\vec{b}| = 1 \) and \( |\vec{a} \times \vec{b}|^2 = k^2 - (\vec{a} \cdot \vec{b})^2 \), then the values of \( k \) and \( \theta \) are:

• A. \( k=1, \theta=45^\circ \)
• B. \( k=7, \theta=60^\circ \)
• C. \( k=49, \theta=90^\circ \)
• D. \( k=7 \) and \( \theta \) is arbitrary

Hint:

Whenever an angle \( \theta \) disappears from the final simplified equation, it means \( \theta \) is arbitrary and not uniquely determined by the given conditions.

Answer 1

The Correct Option is D

Step 1: Understand the given information.
We are given:
\( |\vec{a}| = 7 \)
\( |\vec{b}| = 1 \)
\( |\vec{a} \times \vec{b}|^2 = k^2 - (\vec{a} \cdot \vec{b})^2 \)

Step 2: Express known quantities.
We know: \[ |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta = 7\sin\theta \] thus: \[ |\vec{a} \times \vec{b}|^2 = 49\sin^2\theta \] Also: \[ \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta = 7\cos\theta \quad \Rightarrow \quad (\vec{a} \cdot \vec{b})^2 = 49\cos^2\theta \] Thus, the given condition becomes: \[ 49\sin^2\theta = k^2 - 49\cos^2\theta \]
Step 3: Simplify.
Use identity: \[ \sin^2\theta + \cos^2\theta = 1 \quad \Rightarrow \quad \sin^2\theta = 1 - \cos^2\theta \] Thus: \[ 49(1 - \cos^2\theta) = k^2 - 49\cos^2\theta \] Expanding: \[ 49 - 49\cos^2\theta = k^2 - 49\cos^2\theta \] Now cancel \( -49\cos^2\theta \) from both sides: \[ 49 = k^2 \quad \Rightarrow \quad k = 7 \quad (\text{taking positive value}) \]
Step 4: Analyze \( \theta \).
Notice carefully — \( \theta \) does not appear anymore after solving for \( k \).
\( \theta \) is not determined uniquely.
\( \theta \) can be arbitrary (any value).
Thus, \( \theta \) can take any value and \( k = 7 \).
Step 5: Final Conclusion.
Thus, \( k=7 \) and \( \theta \) is arbitrary.