Pearson Edexcel IAL Mathematics P4.7.7 scalar product

Practise scalar products to find angles, perpendicular conditions and exact areas in vector problems involving points, lines and triangles.

Exam points

use a.b = |a||b|cosθ to find exact cosθ or an angle between vectors or lines
apply a perpendicular dot product to form an equation for a constant coordinate
use sinθ from a scalar product to calculate exact triangle or parallelogram area

[Maximum number: 3]

The line $l_1$ has equation
$\mathbf r=\begin{pmatrix}2\\3\\-7\end{pmatrix}+\lambda\begin{pmatrix}1\\2\\2\end{pmatrix}$
where $\lambda$ is a scalar parameter.

The line $l_2$ has equation
$\mathbf r=\begin{pmatrix}2\\3\\-7\end{pmatrix}+\mu\begin{pmatrix}4\\-1\\8\end{pmatrix}$
where $\mu$ is a scalar parameter.

Given that $l_1$ and $l_2$ meet at the point P.

The point Q lies on $l_1$ where $\lambda=6$ .
Given that point R lies on $l_2$ such that triangle QPR is an isosceles triangle with PQ=PR.

Given that the angle between lines $l_1$ and $l_2$ is $\theta$ ,
find the value of $\cos\theta$ , giving the answer as a fully simplified fraction.

$\begin{pmatrix}1\\2\\2\end{pmatrix}\cdot\begin{pmatrix}4\\-1\\8\end{pmatrix}=1\cdot4+2\cdot(-1)+2\cdot8=18$
M1
Attempts $\mathbf a\cdot\mathbf b=|\mathbf a||\mathbf b|\cos\theta$ :
$18=\sqrt{1^2+2^2+2^2}\sqrt{4^2+(-1)^2+8^2}\cos\theta$
dM1
$\cos\theta=\frac23$
A1

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P4.7.7 - Scalar product question 1