Question

The vectors \[ \vec{a}=2\hat{i}+3\hat{j}+6\hat{k} \] and \(\vec{b}\) are collinear and \[ |\vec{b}|=21, \] then \(\vec{b}=\)

• A. \(\pm(2\hat{i}+3\hat{j}+6\hat{k})\)
• B. \(\pm(6\hat{i}+9\hat{j}+18\hat{k})\)
• C. \(\dfrac{21}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})\)
• D. \(\pm21(2\hat{i}+3\hat{j}+6\hat{k})\)

Hint:

If two vectors are collinear, then one is always a scalar multiple of the other: \[ \vec{b}=\lambda\vec{a} \] Use magnitudes to determine the scalar \(\lambda\).

Answer 1

The Correct Option is B

Step 1: Use the condition for collinear vectors.
If two vectors are collinear, then one vector is a scalar multiple of the other.
Therefore, \[ \vec{b}=\lambda \vec{a} \] So, \[ \vec{b}=\lambda(2\hat{i}+3\hat{j}+6\hat{k}) \]

Step 2: Find the magnitude of \(\vec{a}\).
\[ |\vec{a}| = \sqrt{2^2+3^2+6^2} \] \[ = \sqrt{4+9+36} \] \[ = \sqrt{49} \] \[ =7 \]

Step 3: Use the magnitude relation.
Since \[ \vec{b}=\lambda\vec{a}, \] we have \[ |\vec{b}|=|\lambda||\vec{a}| \] Given, \[ |\vec{b}|=21 \] Thus, \[ 21=|\lambda|\times 7 \]

Step 4: Solve for \(|\lambda|\).
\[ |\lambda|=\frac{21}{7} \] \[ |\lambda|=3 \] Therefore, \[ \lambda=\pm 3 \]

Step 5: Find \(\vec{b}\).
Substituting, \[ \vec{b}=\pm3(2\hat{i}+3\hat{j}+6\hat{k}) \] \[ =\pm(6\hat{i}+9\hat{j}+18\hat{k}) \]

Step 6: Verify the magnitude.
\[ |(6,9,18)| = \sqrt{36+81+324} \] \[ = \sqrt{441} \] \[ =21 \] Hence, the condition is satisfied.

Step 7: Final conclusion.
Therefore, \[ \boxed{\pm(6\hat{i}+9\hat{j}+18\hat{k})} \]