The angle ' θ ' between the vectors {a} = {i} - {j} + {k} and {b} = {i} + {j} - {k} is \_\_\_\_

Correct Answer is : cos⁡−1(−13)^{-1}≤ft(-1/3)cos−1(−31)

Correct Answer is : cos⁡−1(−13)^{-1}≤ft(-1/3)cos−1(−31)

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GUJCET 2024 Math Paper (04)

Question : 2 of 40

Marks: +1, -0

The angle '

\theta

\vec{a} = \hat{i} - \hat{j} + \hat{k}

\vec{b} = \hat{i} + \hat{j} - \hat{k}

\_\_\_\_

Solution:

Concept:

Use the dot product formula to find the angle between two vectors.

Explanation:

Given

\vec{a}=\hat{i}-\hat{j}+\hat{k}

and

\vec{b}=\hat{i}+\hat{j}-\hat{k}

Dot product:

\vec{a}\cdot\vec{b}=1-1-1=-1

Magnitudes:

|\vec{a}|=\sqrt{1^2+(-1)^2+1^2}=\sqrt{3}

|\vec{b}|=\sqrt{1^2+1^2+(-1)^2}=\sqrt{3}

Thus,

\cos\theta=\frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}=-\frac{1}{3}

So,

\theta=\cos^{-1}\left(-\frac{1}{3}\right)

Answer:

\cos^{-1}\left(-\frac{1}{3}\right)

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