Question

Three lines, $L_{1}, L_{2}$ and $L_{3},$ have vector equations

$r = 16 i - 4 j - 6 k + λ (- 12 i + 4 j + 3 k)$
$r = 16 i + 28 j + 15 k + μ (8 i + 8 j + 5 k)$
$r = i + 9 j + 3 k + v (4 i - 12 j - 8 k)$

The $3$ points of intersection of these lines form an acute-angled triangle. For this triangle, find the size of each of the interior angles

Accepted Answer

Given that: Three lines, $L_{1}, L_{2}$ and $L_{3},$ have vector equations

$r = 16 i - 4 j - 6 k + λ (- 12 i + 4 j + 3 k)$
$r = 16 i + 28 j + 15 k + μ (8 i + 8 j + 5 k)$
$r = i + 9 j + 3 k + v (4 i - 12 j - 8 k)$

We have to find the size of each of the interior angles

We have, $r = 16 i - 4 j - 6 k + λ (- 12 i + 4 j + 3 k)$ and $r = 16 i + 28 j + 15 k + μ (8 i + 8 j + 5 k)$

Now equate the components

$x = - 12 λ + 16 x = 8 μ + 16 . . . . . (1)$

$y = 4 λ - 4 y = 8 μ + 28 . . . . . (2)$

$z = 3 λ - 6 z = 5 μ + 15 . . . . . (3)$

Attempt to solve these equations to find the value of $λ$ and of $μ$ that should give a consistent point:

From $(2)$ : $λ = 2 μ + 8$

Substitute into $(3)$ : $3 (2 μ + 8) - 6 = 5 μ + 15$

$\Rightarrow 3 (2 μ + 8) = 5 μ + 21$

$\Rightarrow μ + 24 = 21$

$\Rightarrow μ = - 3$

Then, $λ = 2 (- 3) + 8 = 2$

Check:

$λ = 2$ $μ = - 3$

$x = - 8$ $x = - 8$

$y = 4$ $y = 4$

$z = 0$ $z = 0$

The value of $λ$ and $μ$ give a consistent point.

So there is a point of intersection and the lines are intersecting.

Therefore, the position vector of vertex is $- 8 i + 4 j + 0 k$ .

Similarly, the position vector of vertices are $4 i - 3 k, 12 j + 5 k$

The position vectors of sides are $- 12 i + 4 j + 3 k, 8 i + 8 j + 5 k, - 4 i + 12 j + 8 k$

Now, the angle between $- 12 i + 4 j + 3 k$ and $8 i + 8 j + 5 k$ is $θ = \cos^{- 1} (\frac{- 12 (8) + 4 \times 8 + 3 \times 5}{\sqrt{{(- 12)}^{2} + 4^{2} + 3^{2}} \sqrt{8^{2} + 8^{2} + 5^{2}}})$

$\Rightarrow θ = \cos^{- 1} (\frac{- 96 + 32 + 15}{13 \sqrt{153}})$

$\Rightarrow θ = \cos^{- 1} (\frac{- 49}{13 \sqrt{153}}) = 72.3 °$ (Take acute angle)

Similarly, other two angles are $55.8 °, 51.9 °$

Hence, the size of each of the interior angles are $55.8 °, 72.3 °, 51.9 °$ .

Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Vectors, Exercise 9: EXERCISE 9E

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Vectors, Exercise 9: EXERCISE 9E

Questions from Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Vectors, Exercise 9: EXERCISE 9E with Hints & Solutions

Learn and Prepare for any exam you want !