Question

If A vector is equals to 3 iCAP + 4 j cap and B vector equals to icap + J cap + 2 K cap then find out unit vector along a vector A+ Bvector

Answer 1

(4iCAP+5Jcap+2kcap)/root45

Answer 2

Answer:

The unit vector along a A+B is $\hat{C}=\dfrac{4}{\sqrt{45}}\hat{i}+\dfrac{5}{\sqrt{45}}\hat{j}+\dfrac{2}{\sqrt{45}}\hat{k}$

Explanation:

Given that,

Vector $\vec{A} =3\hat{i}+4\hat{j}$

Vector $\vec{B} =\hat{i}+\hat{j}+2\hat{k}$

The vector A+B is

$\vec{A+B}=\vec{C}=3\hat{i}+4\hat{j}+\hat{i}+\hat{j}+2\hat{k}$

$\vec{C}=4\hat{i}+5\hat{j}+2\hat{k}$

Now, the magnitude of A+B

$|\vec{C}|=\sqrt{4^2+5^2+2^2}$

$|\vec{C}|=\sqrt{45}$

The unit vector along a A+B is

$\hat{C}=\dfrac{\vec{C}}{|\vec{C}|}$

$\hat{C}=\dfrac{4\hat{i}+5\hat{j}+2\hat{k}}{\sqrt{45}}$

$\hat{C}=\dfrac{4}{\sqrt{45}}\hat{i}+\dfrac{5}{\sqrt{45}}\hat{j}+\dfrac{2}{\sqrt{45}}\hat{k}$

Hence, The unit vector along a A+B is $\hat{C}=\dfrac{4}{\sqrt{45}}\hat{i}+\dfrac{5}{\sqrt{45}}\hat{j}+\dfrac{2}{\sqrt{45}}\hat{k}$