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 \(\cos\alpha+\cos^2\beta+7\cos^3\gamma=\)
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 \(\cos\alpha+\cos^2\beta+7\cos^3\gamma=\)
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Direction cosines are obtained by dividing vector components by the vector magnitude.
Updated On: Jun 17, 2026
- \(1\)
- \(\dfrac{65}{49}\)
- \(2\)
- \(-\dfrac{7}{49}\)
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The Correct Option is B
Solution and Explanation
Concept:
Direction cosines of a vector are obtained by dividing each component by the magnitude.
Step 1: Find the magnitude of the vector.
Given vector: \[ \vec a=3\hat i-6\hat j+2\hat k \] Magnitude: \[ |\vec a| = \sqrt{3^2+(-6)^2+2^2} \] \[ = \sqrt{9+36+4} \] \[ = \sqrt{49}=7 \]
Step 2: Find the direction cosines.
\[ \cos\alpha=\frac37 \] \[ \cos\beta=-\frac67 \] \[ \cos\gamma=\frac27 \]
Step 3: Substitute into the expression.
\[ \cos\alpha+\cos^2\beta+7\cos^3\gamma \] \[ = \frac37+\left(-\frac67\right)^2 +7\left(\frac27\right)^3 \] \[ = \frac37+\frac{36}{49} +\frac{56}{343} \] Taking LCM \(343\), \[ = \frac{147+252+56}{343} \] \[ = \frac{455}{343} \] \[ = \frac{65}{49} \] Hence, \[ \boxed{\frac{65}{49}} \]
Step 1: Find the magnitude of the vector.
Given vector: \[ \vec a=3\hat i-6\hat j+2\hat k \] Magnitude: \[ |\vec a| = \sqrt{3^2+(-6)^2+2^2} \] \[ = \sqrt{9+36+4} \] \[ = \sqrt{49}=7 \]
Step 2: Find the direction cosines.
\[ \cos\alpha=\frac37 \] \[ \cos\beta=-\frac67 \] \[ \cos\gamma=\frac27 \]
Step 3: Substitute into the expression.
\[ \cos\alpha+\cos^2\beta+7\cos^3\gamma \] \[ = \frac37+\left(-\frac67\right)^2 +7\left(\frac27\right)^3 \] \[ = \frac37+\frac{36}{49} +\frac{56}{343} \] Taking LCM \(343\), \[ = \frac{147+252+56}{343} \] \[ = \frac{455}{343} \] \[ = \frac{65}{49} \] Hence, \[ \boxed{\frac{65}{49}} \]
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