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Let $\vec{u}=\hat{i}+\hat{j},\; \; \vec{v}=\hat{i}-\hat{j}\; and\; \vec{w}=\hat{i}+2\hat{j}+3\hat{k}.$ If $\hat{n}$ is a unit vector such that $\vec{u}\cdot \hat{n}=0$ and $\vec{v}\cdot \hat{n}=0,$ then $\left |\vec{w}\cdot \vec{n} \right |$ is equal to

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As we learnt in

Vector Product of two vectors(cross product) -

If $\vec{a}$ and $\vec{b}$ are two vectors and $\Theta$ is the angle between them , then $\vec{a}\times \vec{b}=\left |\vec{a} \left | \right |\vec{b} \right |Sin\Theta \hat{n}$

- wherein

$\hat{n}$ is unit vector perpendicular to both $\vec{a} \: and \: \vec{b}$

Scalar Product of two vectors (dot product) -

$\vec{a}\vec{b}=\left | a \right |\left | b \right |Cos\theta$

- wherein

$\Theta$ is the angle between the vectors $\vec{a}\: and\:\vec{b}$

Since $\vec{n}$ is perpendicular to $\vec{u} \ and \ \vec{v}$ $\vec{u} \times \vec{v} =\begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & 1 & 0 \\ 1 & -1 & 0 \end{vmatrix} = -2 \hat k$

Unit vector of $\vec{u} \times \vec{v} =\hat{k}$

Now, $\begin{vmatrix} \hat w. \hat n \end{vmatrix} =3$

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