1.The position vectors of vertices of a △ABC\triangle ABC△ABC are 4i^−2j^4\hat{i} - 2\hat{j}4i^−2j^, i^+4j^−3k^\hat{i} + 4\hat{j} - 3\hat{k}i^+4j^−3k^ and −i^+5j^+k^-\hat{i} + 5\hat{j} + \hat{k}−i^+5j^+k^ respectively, then ∠ABC\angle ABC∠ABC is equal toa.π6\frac{\pi}{6}6πb.π4\frac{\pi}{4}4πc.π3\frac{\pi}{3}3πd.π2\frac{\pi}{2}2πLogin to continueOnly logged in users canattempt or see the solution.
