Question

Vector which is perpendicular to $a\,\cos \,\theta \,\hat{i}+b\,\sin \,\theta \,\,\hat{j}$ is

• A. $b\,\sin \,\theta \,\hat{i}-a\,\cos \,\theta \,\hat{j}$
• B. $\frac{1}{a}\,\sin \,\theta \,\hat{i}-a\,\cos \,\theta \,\hat{j}$
• C. $5\hat{k}$
• D. All of the above

Answer 1

Answer: (D)

All of the above

Answer 2

From definition of dot product of vectors, we have
$x\,.\,y=xy\,\,\cos \,\,\theta$
When $\theta ={{90}^{o}},\,\,\,\cos \,\,{{90}^{o}}=0$
$\therefore$ $x\,.\,y=0$
Given, $x=a\,\cos \,\theta \,\hat{i}+b\,\sin \,\theta \,\hat{j}$
$y=b\,sin\,\,\theta \,\,\hat{i}-a\,\cos \,\,\theta \,\hat{j}$
$x\,.\,y=(a\,\cos \,\,\theta \,\,\hat{i}+b\,sin\,\,\theta \,\hat{j})$
$(b\,sin\,\,\theta \,\,\hat{i}-a\,\cos \,\,\theta \,\hat{j})$
$x\,.\,y=ab\,\,\sin \,\theta \,\,\cos \,\theta \,-\,ab\,\,\,\sin \,\theta \,\,\cos \,\theta \,=0$
Hence, vectors are perpendicular.
Similarly for option and also
$x.y=0$