Question

The direction cosines of the vector \[ \vec a=a_1\hat i+a_2\hat j+a_3\hat k \] are

• A. \(a_1,a_2,a_3\)
• B. \(\dfrac{a_1}{|\vec a|},\dfrac{a_2}{|\vec a|},\dfrac{a_3}{|\vec a|}\)
• C. \(\dfrac{a_1^2}{|\vec a|},\dfrac{a_2^2}{|\vec a|},\dfrac{a_3^2}{|\vec a|}\)
• D. \(\cos a_1,\cos a_2,\cos a_3\)

Hint:

For a vector \[ \vec a=a_1\hat i+a_2\hat j+a_3\hat k \] Direction cosines are obtained by dividing each component by the magnitude: \[ (l,m,n) = \left( \frac{a_1}{|\vec a|}, \frac{a_2}{|\vec a|}, \frac{a_3}{|\vec a|} \right) \]

Answer 1

The Correct Option is B

Concept: If a vector \[ \vec a=a_1\hat i+a_2\hat j+a_3\hat k \] makes angles \(\alpha,\beta,\gamma\) with the positive \(x,y,z\)-axes respectively, then its direction cosines are \[ l=\cos\alpha,\quad m=\cos\beta,\quad n=\cos\gamma \] and \[\begin{aligned} l=\frac{a_1}{|\vec a|}, \qquad m=\frac{a_2}{|\vec a|}, \qquad n=\frac{a_3}{|\vec a|} \end{aligned}\]

Step 1: Find the magnitude of the vector. \[\begin{aligned} |\vec a| = \sqrt{a_1^2+a_2^2+a_3^2} \end{aligned}\]

Step 2: Write the direction cosines. \[\begin{aligned} \left( \frac{a_1}{|\vec a|}, \frac{a_2}{|\vec a|}, \frac{a_3}{|\vec a|} \right) \end{aligned}\] \[\begin{aligned} \boxed{ \frac{a_1}{|\vec a|}, \frac{a_2}{|\vec a|}, \frac{a_3}{|\vec a|} } \end{aligned}\] Hence, option \(\mathbf{(B)}\) is correct.