The direction cosines of the unit vector perpendicular to the plane vector r. (2i + 3j -6k) + 1 = 0 and passing through the origin are

Question

The direction cosines of the unit vector perpendicular to the plane \(\vec r . (2\hat i + 3 \hat j - 6\hat k) + 1 =0\) and passing through the origin are

(A) \(\frac{1}{\sqrt3}, \frac{1}{\sqrt3},\frac{-1}{\sqrt3}\)

(B) \(\frac{-2}{\sqrt7}, \frac{-3}{\sqrt7},\frac{6}{\sqrt7}\)

(C) \(\frac{-2}{7}, \frac{-3}{7}, \frac{6}{7}\)

(D) 1, 0, 0

Answer 1

Correct option is (C) \(\frac {-2}{7}, \frac {-3}7, \frac {6}7\)

\(\vec r. (2\hat i + 3\hat j - 6\hat k) + 1 = 0\)

\(\vec r. (2\hat i + 3\hat j - 6\hat k) = -1\)

\(-2\hat i - 3\hat j + 6\hat k = 1\)

\(\Rightarrow\frac {-2\hat i - 3\hat j + 6\hat k }{\sqrt{2^2 + 3^2 + 6^2}} =\frac {-2\hat i - 3\hat j + 6\hat k }{\sqrt{4 + 9 +36}}\)

\( =\frac {-2\hat i - 3\hat j + 6\hat k }{\sqrt{49}}\)

\( =\frac {-2\hat i - 3\hat j + 6\hat k }{7}\)

\(= \frac {-2\hat i}7-\frac {3\hat j}7 + \frac {6\hat k}7\)

Hence, direction cosines are \(\frac {-2}{7}, \frac {-3}7, \frac {6}7\).

Answer 2

Correct option is (C) \(\frac{-2}{7}, \frac{-3}{7}, \frac{6}{7}\)