If \( |\vec{a}|=3 \) and \( |\vec{b}|=4 \), then a value of λ for which \( \vec{a}+\lambda\vec{b} \) and \( \vec{a}-\lambda\vec{b} \) are perpendiculars is.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 3/4
Two vectors are perpendicular when their dot product is zero:
\( \displaystyle (\vec{a}+\lambda\vec{b})\cdot(\vec{a}-\lambda\vec{b})=0 \)
\( \displaystyle |\vec{a}|^2-\lambda^2|\vec{b}|^2=0 \)
The cross terms \( -\lambda\,\vec{a}\cdot\vec{b}+\lambda\,\vec{b}\cdot\vec{a} \) cancel.
\( \displaystyle 9-16\lambda^2=0 \)
\( \displaystyle \lambda^2=\frac{9}{16} \)
\( \displaystyle \lambda=\pm\frac{3}{4} \)
So a value of λ is 3/4.
Why the other options are incorrect
Option (1) 9/16 — Incorrect. 9/16 is λ2; the square root has not been taken.
Option (3) 3/2 — Incorrect. With λ = 3/2, \( 9-16\lambda^2=9-36=-27\ne 0 \).
Option (4) 4/3 — Incorrect. 4/3 is the ratio inverted; with λ = 4/3, \( 9-16\lambda^2\ne 0 \).