A unit vector in the xy-plane which is perpendicular to the vector \(3\hat{\imath}-4\hat{\jmath}\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{1}{5}(4\hat{\imath}+3\hat{\jmath})\)
Let the required vector be \(x\hat{\imath}+y\hat{\jmath}\) (it lies in the xy-plane, so it has no \(\hat{k}\) part).
Since it is perpendicular to \(3\hat{\imath}-4\hat{\jmath}\), the scalar product is zero:
\(3x-4y = 0\)
\( \displaystyle y = \frac{3x}{4} \)
Since it is a unit vector:
\( \displaystyle x^2+\frac{9x^2}{16} = 1 \)
\( \displaystyle x = \pm\frac{4}{5}, \quad y = \pm\frac{3}{5} \)
So the unit vectors are \(\pm\displaystyle\frac{1}{5}(4\hat{\imath}+3\hat{\jmath})\). Of these, \(\displaystyle\frac{1}{5}(4\hat{\imath}+3\hat{\jmath})\) is among the options.
Check: \((3)(4)+(-4)(3) = 0\).