# Find the scalar value and calculate perpendicular vector

Stuck on this problem for my class for a bit, I'm not sure exactly how to solve it. If someone could point me in the right direction I would really appreciate it.

What I have so far

Given the points: A = ( 3, -4, 2) B = (-1, -3, 5)

and the vector:

n = 2i + 5j - k

Find the displacement vector of AB = (-4, 1, 3)

Calculate the dot product (scalar product) OA . n = -16

Let THETA be the angle between the vectors OA and n. THETA is greater than 90 degrees.

Now here is where I'm stuck.

The vector OA is to be expressed as a sum OA = kn + b, where k is a scalar and b is a vector perpendicular to n. Determine the value of k in this expression.

I know the answer is staring me in the face but I'm feeling very math-illiterate.

Consider the plane with normal $\mathbf n$ and containing the point $A$. The equation of this plane, as you have worked out earlier, is $$\mathbf r\cdot\mathbf n=-16$$ The asked-for decomposition requires the perpendicular component to reach from the origin to a point on this plane – in other words, $k\mathbf n\cdot\mathbf n=-16$. Solving, we get $30k=-16$ and hence $k=-\frac8{15}$. Then $$\mathbf b=\vec{OA}-k\mathbf n=(3,-4,2)+\tfrac8{15}(2,5,-1)=(\tfrac{61}{15},-\tfrac43,\tfrac{22}{15})$$ and we can verify that $\mathbf b\cdot\mathbf n=0$, i.e. they are perpendicular as stipulated in the question.
• There's an old adage that says "you learn maths by doing maths". Just keep on doing questions, and you will understand the structures they deal with (and how to solve them). I merely substituted $k\mathbf n$, the perpendicular component, into the formula for the plane I showed above. It turns out that the general equation for any plane can be written (vector)•(normal to plane)=constant. – Parcly Taxel Sep 8 '16 at 5:46
Since $b\cdot n=0$, take the dot product with n of both sides of $OA=kn+b$ and then solve for $k$.
Then use $b=OA-kn$.