2022TrialShort answer

Q13 (c) (ii)

3 marks

Question

The equation of line $l_2$ is
$\frac{x+3}{k}=\frac{4-y}{6}=z-1.$
Find the value of $k$ for which $l_1$ and $l_2$ intersect.

Try the question before revealing the answer.

Solution

Line 1

x=2+\lambda\qquad\text{(I)}

y=-1-\lambda\qquad\text{(II)}

z=4-\lambda\qquad\text{(III)}

Line 2

\frac{x+3}{k}=\frac{4-y}{6}\qquad\text{(A)}

\frac{4-y}{6}=z-1\qquad\text{(B)}

Substitute Equations II and III into equation B:

\frac{4-(-1-\lambda)}{6}=4-\lambda-1

5+\lambda=18-6\lambda

\lambda=\frac{13}{7}

Substitute equations I and II into equation A:

\frac{2+\lambda+3}{k}=4-\lambda-1

5+\lambda=k(3-\lambda)

k=\frac{5+\lambda}{3-\lambda}

Substituting the value for $\lambda$ :

k=\frac{5+\frac{13}{7}}{3-\frac{13}{7}}=6

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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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