Question

If $(2, - 1,2)$ and $(K,3,5)$are the triads of direction ratios of two lines and the angle between them is ${45^ \circ }$, then a value of $K$ is
A. $2$
B. $3$
C. $4$
D. $6$

Answer 1

Hint: Apply the formula to find angle between two points:
Angle between two points: If \[\theta \] is the angle between two lines whose direction ratios are proportional to \[{a_1},{\text{ }}{b_1},{\text{ }}{c_1}\] and \[{a_2},{\text{ }}{b_2},{\text{ }}{c_2}\] respectively, then the angle \[\theta \] between them is given by,
\[\cos \theta = \dfrac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {a_1^2 + b_1^2 + c_1^2} \sqrt {a_2^2 + b_2^2 + c_2^2} }}\]
Here, substitute\[{a_1} = 2\],\[{b_1} = - 1\], \[{c_1} = 2\] , and \[{a_2} = K\],\[{b_2} = 3\],\[{c_2} = 5\]and angle $\theta = {45^ \circ }$ and solve for \[K\].
Here, $\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$.

Complete step-by-step answer:
If the given triads $(2, - 1,2)$ and $(K,3,5)$are the directions. The angle between two lines is ${45^ \circ }$.
Angle between two points: If \[\theta \] is the angle between two lines whose direction ratios are proportional to \[{a_1},{\text{ }}{b_1},{\text{ }}{c_1}\] and \[{a_2},{\text{ }}{b_2},{\text{ }}{c_2}\]respectively, then the angle \[\theta \] between them is given by,
\[\cos \theta = \dfrac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {a_1^2 + b_1^2 + c_1^2} \sqrt {a_2^2 + b_2^2 + c_2^2} }} \ldots \ldots (1)\]
Substitute\[{a_1} = 2\],\[{b_1} = - 1\], \[{c_1} = 2\] , and \[{a_2} = K\],\[{b_2} = 3\],\[{c_2} = 5\]and angle $\theta = {45^ \circ }$ into the equation $(1)$.
\[\cos {45^ \circ } = \dfrac{{2 \times K + ( - 1) \times 3 + 2 \times 5}}{{\sqrt {{2^2} + {{( - 1)}^2} + {2^2}} \sqrt {{K^2} + {3^2} + {5^2}} }}\]
Simplify the equation we get,
\[\cos {45^ \circ } = \dfrac{{2K - 3 + 10}}{{\sqrt {4 + 1 + 4} \sqrt {{K^2} + 9 + 25} }}\]
We know trigonometric value$\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$, substitute into the equation and simplify,
\[\dfrac{1}{{\sqrt 2 }} = \dfrac{{2K + 7}}{{\sqrt 9 \sqrt {{K^2} + 34} }}\]
\[ \Rightarrow \dfrac{1}{{\sqrt 2 }} = \dfrac{{2K + 7}}{{3\sqrt {{K^2} + 34} }}\]
Squaring both the sides of the equation,
\[ \Rightarrow \dfrac{1}{2} = \dfrac{{{{(2K + 7)}^2}}}{{9({K^2} + 34)}}\]
\[ \Rightarrow 9({K^2} + 34) = 2{(2K + 7)^2}\]
\[ \Rightarrow 9{K^2} + 306 = 2(4{K^2} + 28K + 49)\]
Simplify for $K$,
\[ \Rightarrow 9{K^2} - 8{K^2} - 56K + 306 - 98 = 0\]
\[ \Rightarrow {K^2} - 56K + 208 = 0\]
 Find the factors of quadratic equation,
\[ \Rightarrow (K - 52)(K - 4) = 0\]
The value of $K$ is $4,52$ .

The correct Answer: C. $4$

Note:
We can solve the quadratic equation by the following formula;
If $a{x^2} + bx + c = 0$ is the quadratic equation, then $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$,
where $a$ and $b$ are the coefficients of ${x^2}$, $x$ and c is the constant.