Question

Let, \(a=i-j+2k\). Then the vector in the direction of a with magnitude \(5\) units is ?

• A. \(5i-5j+10k\)
• B. \(-5i-5j+10k\)
• C. \(\dfrac{1}{√6}[5i-5j+10k]\)
• D. \(\dfrac{1}{√6}[10i-5j+5k]\)

Answer 1

The Correct Option is C

Given: \(a  =i-j+2k\)

Then, first find magnitude of \(a\)

\(|a|=√(1^{2}+(-1)^{2}+2^{2})\)

      \(=√6\)

Then the unit vector in the direction of a is 

    \(=\)  \(\dfrac{1}{√6}[i-j+2k]\)

as per the question the vector having a magnitude of 5 

    \(=\)\(\dfrac{1}{√6}[5i-5j+10k]\) 

Answer 2

Step 1: Find the unit vector in the direction of \(\vec{a}\). Given \(\vec{a} = \vec{i} - \vec{j} + 2\vec{k}\), first compute its magnitude: \[ |\vec{a}| = \sqrt{1^2 + (-1)^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \] The unit vector \(\hat{a}\) is: \[ \hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{\vec{i} - \vec{j} + 2\vec{k}}{\sqrt{6}} \]

Step 2: Scale the unit vector to have magnitude 5. Multiply \(\hat{a}\) by 5 to get the desired vector: \[ 5\hat{a} = \frac{5(\vec{i} - \vec{j} + 2\vec{k})}{\sqrt{6}} = \frac{5\vec{i} - 5\vec{j} + 10\vec{k}}{\sqrt{6}} \]

Step 3: Compare with the given options. The result matches option (C): \[ \boxed{C} \quad \left( \frac{1}{\sqrt{6}}(5\vec{i} - 5\vec{j} + 10\vec{k}) \right) \]