Let a=3i+j-k and c=2i-3j+3k.

Question

Let $\vec{a}=3\hat{i}+\hat{j}-\hat{k}$ and $\vec{c}=2\hat{i}-3\hat{j}+3\hat{k}$. If $\vec{b}$ is a vector such that $\vec{a}=\vec{b}	imes \vec{c}$ and $|\vec{b}{|}^{2}=50$, then $|72-|\vec{b}+{\vec{c}|}^{2}|$ is equal to \_\_\_\_\_\_\_\_\_\_.

Answer: 66

Accepted Answer

Given that $\vec{a}=\vec{b}	imes \vec{c}$, we can find the magnitudes of $\vec{a}$ and $\vec{c}$:

$|\vec{a}|=\sqrt{{3}^{2}+{1}^{2}+(-1{)}^{2}}=\sqrt{11}$

$|\vec{c}|=\sqrt{{2}^{2}+(-3{)}^{2}+{3}^{2}}=\sqrt{22}$

We know that the magnitude of the cross product of two vectors is equal to the product of the magnitudes of the vectors and the sine of the angle between them :

$|\vec{a}|=|\vec{b}	imes \vec{c}|=|\vec{b}||\vec{c}|\sin 	heta $

Plugging in the known values :

$\sqrt{11}=\sqrt{50}\sqrt{22}\sin 	heta $

Solving for the sine of the angle between the vectors :

$\sin 	heta =\frac{1}{10}$

Now we can find $|\vec{b}+\vec{c}{|}^{2}$ using the formula :

$|\vec{b}+\vec{c}{|}^{2}=|\vec{b}{|}^{2}+|\vec{c}{|}^{2}+2\vec{b}\cdot \vec{c}$

We have the dot product $\vec{b}\cdot \vec{c}=|\vec{b}||\vec{c}|\cos 	heta $, and we can use the relationship between sine and cosine: $\cos 	heta =\sqrt{1-{\sin }^{2}	heta }=\frac{\sqrt{99}}{10}$.

Substitute the values into the formula :

$|\vec{b}+\vec{c}{|}^{2}=50+22+2\sqrt{50}\sqrt{22}\frac{\sqrt{99}}{10}=72+66$

Finally, we need to find the absolute value of the difference :

$|72-|\vec{b}+\vec{c}{|}^{2}|=|72-(72+66)|=66$

Solutions · 1