# Let k be a constant. The points P=(4,-2,2), Q=(-4,1,6) and R=(-1,-1,k) in \mathbb{R}^3 are the...

## Question:

Let {eq}k {/eq} be a constant. The points {eq}P=(4,-2,2), Q=(-4,1,6) {/eq} and {eq}R=(-1,-1,k) {/eq} in {eq}\mathbb{R}^3 {/eq} are the vertices of a right training with right angle at the vertex. What is the value of {eq}k {/eq}?

## Vector:

A vector is a quantity that has magnitude as well as direction.

We know that two vectors {eq}\left ( u,v,w \right )\,\,\left ( a,b,c \right ) {/eq} are perpendicular to each other if {eq}\frac{au+vb+wc}{\left \| \left ( u,v,w \right ) \right \|\left \| \left ( a,b,c \right ) \right \|}=\cos \theta {/eq}

Take {eq}\left ( u,v,w \right )=(-4,1,6)\,\,\left ( a,b,c \right )=(-1,-1,k) {/eq}

Take {eq}\theta =90^{\circ} {/eq} to find the value of k.

Let {eq}\left ( u,v,w \right )=(-4,1,6)\,\,\left ( a,b,c \right )=(-1,-1,k) {/eq}

{eq}au+vb+wc=-4(-1)+1(-1)+6(k)=4-1+6k {/eq}

We know that {eq}\frac{au+vb+wc}{\left \| \left ( u,v,w \right ) \right \|\left \| \left ( a,b,c \right ) \right \|}=\cos \theta {/eq}

{eq}\frac{4-1+6k}{\left \| (-4,1,6) \right \|\left \| (-1,-1,k) \right \|}=\cos \theta =\cos90^{\circ}\\ 3+6k=0\,\,\,\,\left \{ \because \cos90^{\circ}=0 \right \}\\ 6k=-3\\ k=\frac{-3}{6}\\ k=\frac{-1}{2} {/eq}