Question

If \(\vec{a}=2\hat{i}+\hat{j}-8\hat{k}\) and \(\vec{b}=\hat{i}+3\hat{j}-4\hat{k}\), then the magnitude of \(\vec{a}+\vec{b}\) is

• A. \(13\)
• B. \(\frac{13}{3}\)
• C. \(\frac{3}{13}\)
• D. \(\frac{4}{13}\)

Hint:

For vector addition, add the corresponding \(\hat{i},\hat{j},\hat{k}\) components first, then use magnitude formula \(\sqrt{x^2+y^2+z^2}\).

Answer 1

The Correct Option is A

Concept:
To find the magnitude of the sum of two vectors, first add the corresponding components of the vectors. If: \[ \vec{v}=x\hat{i}+y\hat{j}+z\hat{k} \] then its magnitude is: \[ |\vec{v}|=\sqrt{x^2+y^2+z^2} \]

Step 1: Write the given vectors.
\[ \vec{a}=2\hat{i}+\hat{j}-8\hat{k} \] \[ \vec{b}=\hat{i}+3\hat{j}-4\hat{k} \]

Step 2: Add the corresponding components.
\[ \vec{a}+\vec{b} = (2\hat{i}+\hat{j}-8\hat{k})+(\hat{i}+3\hat{j}-4\hat{k}) \] Now add the \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) components separately: \[ \hat{i}\text{ component}=2+1=3 \] \[ \hat{j}\text{ component}=1+3=4 \] \[ \hat{k}\text{ component}=-8+(-4)=-12 \] Therefore: \[ \vec{a}+\vec{b}=3\hat{i}+4\hat{j}-12\hat{k} \]

Step 3: Find the magnitude of \(\vec{a}+\vec{b}\).
\[ |\vec{a}+\vec{b}|=\sqrt{3^2+4^2+(-12)^2} \] \[ |\vec{a}+\vec{b}|=\sqrt{9+16+144} \] \[ |\vec{a}+\vec{b}|=\sqrt{169} \] \[ |\vec{a}+\vec{b}|=13 \] Hence, the correct answer is: \[ \boxed{(A)\ 13} \]