Question

The component of vector 2i-3j+2k perpendicular to i+j+k is​

Answer 1

please see the attachment

Answer 2

Answer:

Option 1 - $C=\frac{5}{3}(i-2j+k)$

Explanation:

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

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

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

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

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

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

$\vec{a}\cdot \vec{b}=4-3=1$

Substitute the value in the formula,

$C=2i-3j+2k-\frac{1}{3}\times (i+j+k)$

$C=\frac{6i-9j+6k-i-j-k}{3}$

$C=\frac{5i-10j+5k}{3}$

$C=\frac{5}{3}(i-2j+k)$

Therefore, Option 1 - is correct.

The component of vector a perpendicular to vector b is $C=\frac{5}{3}(i-2j+k)$