First slide

Direction ratios and direction cosines

Question

Direction ratios of two lines are $〈4, k, 5〉$ and $〈2, - 1, 2〉$ . If these lines include an angle $\frac{π}{4}$ , then $k$ can be

Easy

A

1
B

2
C

3
D

4

Solution

If $θ$ is the acute angle between two lines having direction ratios $〈a_{1}, b_{1}, c_{1}〉$ and $〈a_{2}, b_{2}, c_{2}〉$
then $\cos θ = |\frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{{a_{1}}^{2} + {b_{1}}^{2} + {c_{1}}^{2}} \cdot \sqrt{{a_{2}}^{2} + {b_{2}}^{2} + {c_{2}}^{2}}}|$
It implies that
$\begin{matrix} \cos (\frac{π}{4}) = \frac{8 - k + 10}{\sqrt{16 + 25 + k^{2}} \sqrt{4 + 1 + 4}} \\ \frac{1}{\sqrt{2}} = \frac{18 - k}{\sqrt{41 + k^{2}} (3)} \end{matrix}$
Apply square on both sides
$\begin{matrix} \frac{1}{2} = \frac{{(18 - k)}^{2}}{9 (41 + k^{2})} \\ 9 (41 + k^{2}) = 2 (324 + k^{2} - 36 k) \\ 7 k^{2} + 369 = 648 - 72 k \\ 7 k^{2} + 72 k - 279 = 0 \end{matrix}$
Split the middle term to get the solutions for the equation
$\begin{matrix} 7 k^{2} + 93 k - 21 k - 279 = 0 \\ k (7 k + 93) - 3 (7 k + 93) = 0 \\ (k - 3) (7 k + 93) = 0 \end{matrix}$
Therefore, the possible value of $k$ is $3$

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