2022TrialShort answer

Q15 (a) (i)

3 marks

Question

Show by finding it that the only point of intersection between
$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\10\\5\end{pmatrix}+\lambda\begin{pmatrix}2\\-3\\1\end{pmatrix}$
and the sphere
$(x-5)^2+(y+2)^2+(z-3)^2=38$
is $\begin{pmatrix}7\\1\\8\end{pmatrix}$ .

Try the question before revealing the answer.

Solution

Q.15(a)(i)

x=1+2\lambda

y=10-3\lambda

z=5+\lambda

(x-5)^2+(y+2)^2+(z-3)^2=38

(2\lambda-4)^2+(12-3\lambda)^2+(2+\lambda)^2=38

16\lambda^2-84\lambda+126=0

\lambda^2-6\lambda+9=0

(\lambda-3)^2=0

$\therefore$ only 1 point of intersection: $\lambda=3$ .

x=1+2(3)=7

y=10-3(3)=1

z=5+3=8

Point of intersection $=\begin{pmatrix}7\\1\\8\end{pmatrix}$ .

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Syllabus alignment

Vectors

Vectors and vector equations of linesUse coordinates and vector equations for curves, lines, line segments and spheres in two and three dimensions.

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