If \( \vec{a}=5\hat{\imath}-\hat{\jmath}-3\hat{k} \) and \( \vec{b}=\hat{\imath}+3\hat{\jmath}-5\hat{k} \), then which of the following is/are TRUE?
(A) \( \left(\vec{a}+\vec{b}\right)\perp\left(\vec{a}-\vec{b}\right) \)
(B) \( |\vec{a}+\vec{b}|=6 \)
(C) \( |\vec{a}-\vec{b}|=2\sqrt{26} \)
(D) \( \left|\left(\vec{a}+\vec{b}\right)+\left(\vec{a}-\vec{b}\right)\right|=2\sqrt{35} \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (D) only
Adding and subtracting the components:
\( \displaystyle \vec{a}+\vec{b}=6\hat{\imath}+2\hat{\jmath}-8\hat{k} \)
\( \displaystyle \vec{a}-\vec{b}=4\hat{\imath}-4\hat{\jmath}+2\hat{k} \)
\( \displaystyle (\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})=24-8-16=0 \)
\( \displaystyle |\vec{a}+\vec{b}|=\sqrt{36+4+64}=\sqrt{104} \)
\( \displaystyle =2\sqrt{26} \)
\( \displaystyle |\vec{a}-\vec{b}|=\sqrt{16+16+4}=6 \)
The sum of the two vectors is \( 2\vec{a} \):
\( \displaystyle (\vec{a}+\vec{b})+(\vec{a}-\vec{b})=2\vec{a} \)
\( \displaystyle |2\vec{a}|=2\sqrt{25+1+9}=2\sqrt{35} \)
So the dot product is 0, \( |\vec{a}+\vec{b}|=2\sqrt{26} \), \( |\vec{a}-\vec{b}|=6 \) and the magnitude of the sum is \( 2\sqrt{35} \).
Checking the statements
(A) \( \left(\vec{a}+\vec{b}\right)\perp\left(\vec{a}-\vec{b}\right) \) — Correct. Their dot product is zero.
(B) \( |\vec{a}+\vec{b}|=6 \) — Incorrect. \( |\vec{a}+\vec{b}| \) is \( 2\sqrt{26} \); 6 is \( |\vec{a}-\vec{b}| \).
(C) \( |\vec{a}-\vec{b}|=2\sqrt{26} \) — Incorrect. \( |\vec{a}-\vec{b}| \) is 6; \( 2\sqrt{26} \) is \( |\vec{a}+\vec{b}| \).
(D) \( \left|\left(\vec{a}+\vec{b}\right)+\left(\vec{a}-\vec{b}\right)\right|=2\sqrt{35} \) — Correct. The sum is \( 2\vec{a} \), of magnitude \( 2\sqrt{35} \).
Shortcut: \( (\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})=|\vec{a}|^2-|\vec{b}|^2 \). Here \( |\vec{a}|^2=|\vec{b}|^2=35 \), so (A) holds at once.