Question

find the angle between vector A = iCAP + jcap -2 k cap and vector B = i cap + 2 J cap - K cap .



THIS QUESTION IS 11 STANDARD OF VECTOR CHAPTER

Answer 1

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Answer 2

Answer:

The angle between the vector is $\bold{\theta=\cos ^{-1} \frac{5}{6}}$

Explanation:

Given:

Vector $\mathrm{A}=\hat{\imath}+\hat{\jmath}-2 \hat{k}$

Vector $\mathrm{B}=\hat{\imath}+2 \hat{\jmath}-\hat{k}$

To find angle we know the formula,

$\cos \theta=\frac{|\vec{A} \cdot \vec{B}|}{\left[\sqrt{a_{1}^{3}+b_{1}^{3}+c_{2}^{2}}\right]\left[\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}\right]}$

Here $a_{1}=1 b_{1}=1 c_{1}=-2 a_{2}=1 b_{2}=2 c_{2}=1$

Substitute the values in the above formula we get,

$\cos \theta=\frac{[1+2+2]}{[\sqrt{1+1+4}][\sqrt{1+4+1}]}$

By simplifying the above expression we get,

$\cos \theta=\frac{5}{[\sqrt{6}][\sqrt{6}]}$ [We know that[√6][√6]=6]

$\cos \theta=\frac{5}{6}$

To find the angle shift the cos to the RHS

$\theta=\cos ^{-1} \frac{5}{6}$