If direction cosines of a vector of magnitude 3 are (2/3),- (1/3),(2/3) and a 0, then vector is

Question

If direction cosines of a vector of magnitude 3 are (2/3),- (1/3),(2/3) and a > 0, then vector is (A) 2i + j + 2k (B) 2i - j + 2k (C) i -2 j +2 k (D) i+2 j+2 k

Tardigrade Admin · Accepted Answer

Unit vector with direction cosine (2/3),-(1/3), (2/3) is given by hata=(2/3) hat i -(1/3) hat j +(2/3) hat k Thus vector with magnitude 3 and direction cosine as above is given by 3 hata=2 hat i - hat j +2 hat k

Q. If direction cosines of a vector of magnitude $3$ are $\frac{2}{3}, - \frac{1}{3}, \frac{2}{3}$ and $a > 0,$ then vector is

$2 i + j + 2 k$

$2 i - j + 2 k$

$i - 2 j + 2 k$

$i + 2 j + 2 k$

Unit vector with direction cosine $\frac{2}{3}, - \frac{1}{3}, \frac{2}{3}$ is given by
$\overset{a}{^} = \frac{2}{3} \hat{i} - \frac{1}{3} \hat{j} + \frac{2}{3} \hat{k}$
Thus vector with magnitude 3 and direction cosine as above is given by
$3 \overset{a}{^} = 2 \hat{i} - \hat{j} + 2 \hat{k}$

Q. If direction cosines of a vector of magnitude 3 are 32​,−31​,32​ and a>0, then vector is

2i+j+2k

2i−j+2k

i−2j+2k

i+2j+2k