Question

Find a+b,2a+3b,IaI, and Ia-bI. a[1, 2, -3], b=[-2, -1, 5]

Answer 1

Answer: a)
`-1\hat{i}+1\hat{j}-2\hat{k}` b)
`2\hat{i}+\hat{j}+9\hat{k}` c) √14 d) √82.

Explanation:

Since we have given that

`\vec{a}=1\hat{i}+2\hat{j}-3\hat{k}\\\\\vec{b}=-2\hat{i}-\hat{j}+5\hat{k}`

So, we will first find :

1) a + b which is given by

`\vec{a}+\vec{b}=1\hat{i}+2\hat{j}-3\hat{k}-2\hat{i}-\hat{j}+5\hat{k}\\\\\vec{a}+\vec{b}=-1\hat{i}+1\hat{j}-2\hat{k}`

2) 2a+3b is given by

`2(1\hat{i}+2\hat{j}-3\hat{k})+3(-2\hat{i}-\hat{j}+5\hat{k})\\\\=4\hat{i}+4\hat{j}-6\hat{k}-6\hat{i}-3\hat{j}+15\hat{k}\\\\=-2\hat{i}+\hat{j}+9\hat{k}`

3) |a| is given by

`|a|=√(1^2+2^2+(-3)^2)\\\\|a|=√(1+4+9)\\\\|a|=√(14)`

4) |a-b| is given by

`|a-b|=√((1+2)^2+(2+1)^2+(-3-5)^2)\\\\|a-b|=√(3^2+3^2+(-8)^2)\\\\|a-b|=√(9+9+64)\\\\|a-b|=√(82)`

Hence, a)
`-1\hat{i}+1\hat{j}-2\hat{k}` b)
`2\hat{i}+\hat{j}+9\hat{k}` c) √14 d) √82.