Question

If a vector \(3\hat i-6\hat j+2\hat k\) makes angles \(\alpha,\beta,\gamma\) with the positive \(x,y,z\)-axes respectively, then \(\cos\alpha+\cos^2\beta+7\cos^3\gamma=\)

• A. \(1\)
• B. \(\dfrac{65}{49}\)
• C. \(2\)
• D. \(-\dfrac{7}{49}\)

Hint:

Direction cosines are obtained by dividing vector components by the vector magnitude.

Answer 1

The Correct Option is B

Step 1: Find the length of the vector.
For $\vec a=3\hat i-6\hat j+2\hat k$, the magnitude is $\sqrt{3^2+(-6)^2+2^2}=\sqrt{9+36+4}=\sqrt{49}=7$.
Step 2: Get the direction cosines.
Divide each component by $7$: $\cos\alpha=\dfrac37$, $\cos\beta=-\dfrac67$, $\cos\gamma=\dfrac27$.
Step 3: Plug into the expression.
We need $\cos\alpha+\cos^2\beta+7\cos^3\gamma$.
Step 4: Compute each piece.
$\cos\alpha=\dfrac37$, $\cos^2\beta=\dfrac{36}{49}$, and $7\cos^3\gamma=7\cdot\dfrac{8}{343}=\dfrac{56}{343}$.
Step 5: Add over a common base.
Using base $343$: $\dfrac{147}{343}+\dfrac{252}{343}+\dfrac{56}{343}=\dfrac{455}{343}$.
Step 6: Reduce the fraction.
$\dfrac{455}{343}=\dfrac{65}{49}$. \[ \boxed{\frac{65}{49}} \]