Question

find component of a vector equal ICAP + 2 J cap + 3 k cap perpendicular to b vector equal ICAP + J cap + K cap​

Answer 1

here, $\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$

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

we have to find component of vector a perpendicular to vector b

= $\vec{a}-(\vec{a}.\hat{b})\hat{b}$

here, $\hat{b}$ denotes unit vector of b.

so, $\hat{b}=\frac{\vec{b}}{|\vec{b}|}$

= $\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$

now, component of a perpendicular to vector b

= $(\hat{i}+2\hat{j}+3\hat{k})-\left(\hat{i}+2\hat{j}+3\hat{k}.\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})\right)\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$

=$(\hat{i}+2\hat{j}+3\hat{k})-\left(\frac{1+2+3}{\sqrt{3}}\right)\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$

= $(\hat{i}+2\hat{j}+3\hat{k})-2(\hat{i}+\hat{j}+\hat{k})$

= $-\hat{i}+\hat{j}$

Answer 2

Answer:

The component of vector a perpendicular to vector b is

$C=\frac{(\sqrt{3}-6)i+(2\sqrt{3}-6)j+(3\sqrt{3}-6)k}{\sqrt{3}}$

Explanation:

To find : The components of a $\vec{a}=i+2j+3k$ perpendicular to $C=-i+k$

Solution :

The formula to find component of vector a perpendicular to vector b is given by,

$C=\vec{a}-\frac{\vec{a}\cdot \vec{b}}{(|\vec{b}|)^2}\times \vec{b}$

We know, $\vec{a}=i+2j+3k$ and $\vec{b}=i+j+k$

$|\vec{b}|=\sqrt{1^2+1^2+1^2}$

$|\vec{b}|=\sqrt{3}$

$|\vec{b}|^2=3$

$\vec{a}\cdot \vec{b}=1(1)+2(1)+3(1)$

$\vec{a}\cdot \vec{b}=6$

Substitute the value in the formula,

$C=i+2j+3k-\frac{6}{3}\times (i+j+k)$

$C=i+2j+3k-2\times (i+j+k)$

$C=i+2j+3k-2i-2j-2k$

$C=-i+0j+k$

Therefore, The component of vector a perpendicular to vector b is $C=-i+k$