# What is the angle between the following vectors? Vector A = 3i - 2j + k Vector B = 2i + 6j - 6k (A) arccos(12/√266) (B) arccos(-6/√266) (C) arccos(-6/√1064) (D) None of these

## Question

What is the angle between the following vectors?

$$ \begin{array}{l} \vec{A}=3 \hat{i}-2 \hat{j}+\hat{k} \ \vec{B}=2 \hat{i}+6 \hat{j}-6 \hat{k} \end{array} $$

(A) $\cos^{-1}\left(\frac{12}{\sqrt{266}}\right)$ (B) $\cos^{-1}\left(\frac{-6}{\sqrt{266}}\right)$ (C) $\cos^{-1}\left(\frac{-6}{\sqrt{1064}}\right)$ D None of these

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## Answer

The correct option is **B**:

$$ \cos^{-1}\left(\frac{-6}{\sqrt{266}}\right) $$

First, let's find the dot product of the vectors $\vec{A}$ and $\vec{B}$:

$$ \vec{A} \cdot \vec{B} = (3 \times 2) + (-2 \times 6) + (1 \times -6) $$

Calculating the above expression:

$$ = 6 - 12 - 6 = -12 $$

Next, we find the magnitudes of $\vec{A}$ and $\vec{B}$:

$$ |\vec{A}| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14} $$

$$ |\vec{B}| = \sqrt{2^2 + 6^2 + (-6)^2} = \sqrt{4 + 36 + 36} = \sqrt{76} $$

Now, we use the dot product and magnitudes to find $\cos \theta$:

$$ \cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} = \frac{-12}{\sqrt{14} \cdot \sqrt{76}} = \frac{-12}{\sqrt{1064}} $$

Simplifying the expression inside the cosine function:

$$ \therefore \cos \theta = \frac{-6}{\sqrt{266}} $$

Therefore, the angle $\theta$ can be represented as:

$$ \theta = \cos^{-1}\left(\frac{-6}{\sqrt{266}}\right) $$

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