# Mistake in calculations related to vectors and norms

I am rather new to calculus, and am trying to resolve the following question. I have come to an answer, but it is not listed amongst the possible answers, so I would love to know where my reasoning failed...

Vector $$\mathbf u = (1,0,1)$$ is split into two perpendicular vectors where one is in the direction of vector $$\mathbf b = (1,1,2)$$. Calculate the value of $$\left\lVert \mathbf u \right\rVert ^2\left\lVert \mathbf b \right\rVert ^2\left\lVert \mathbf v \right\rVert ^2\left\lVert \mathbf w \right\rVert ^2$$.

$$\left\lVert \mathbf b \right\rVert=\sqrt6$$

$$\left\lVert \mathbf u \right\rVert=\sqrt2$$

$$cosθ = \frac {u·b}{\left\lVert \mathbf u \right\rVert \left\lVert \mathbf b \right\rVert }$$=$$\frac{(1,0,1)(1,1,2)}{\sqrt12}$$=$$\frac {\sqrt 3}{2}$$, where θ refers to the angle between $$\mathbf u$$ and $$\mathbf b$$; therefore θ = 30 degrees.

Therefore $$\left\lVert \mathbf v \right\rVert$$ is equal to $$\sqrt 2 · cos60^\circ$$=$$\frac {1}{2}\sqrt2$$.

Therefore $$\left\lVert \mathbf w \right\rVert$$ is equal to $$\sqrt 2 · sin60^\circ$$=$$\frac {\sqrt6}{2}$$.

However, that leads to a result of 9 when put the norms into the equation mentioned above, and the options given are 1,2,3 and 4.

Can anybody please let me know where I went wrong?

Thank you!

• Your solution looks correct to me. Double-check the problem itself. If, for instance, those norms weren’t squared, then the result would be $3$ instead. – amd Mar 21 at 22:43
• Your solution also looks correct to me, with your result of $9$ matching what I got. – John Omielan Mar 21 at 23:08