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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$.

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

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

$cosθ = \frac {u·b}{\left\lVert \mathbf u \right\rVert \left\lVert \mathbf b \right\rVert }$=$\frac{(1,0,1)(1,1,2)}{\sqrt12}$=$\frac {\sqrt 3}{2}$, 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!

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    $\begingroup$ 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. $\endgroup$ – amd Mar 21 at 22:43
  • $\begingroup$ Your solution also looks correct to me, with your result of $9$ matching what I got. $\endgroup$ – John Omielan Mar 21 at 23:08

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