Question

If \((K,3,5)\), \((2,-1,2)\) are direction ratios of two lines and the angle between them is \(45^\circ\), then a value of \(K\) is:

• A. \(2\)
• B. \(4\)
• C. \(6\)
• D. \(8\)

Hint:

For direction ratios: \[ \cos\theta= \frac{\vec{a}\cdot\vec{b}} {|\vec{a}||\vec{b}|} \] Always simplify the numerator and denominator separately before squaring the equation.

Answer 1

The Correct Option is B

Concept: If two lines have direction ratios: \[ (a_1,b_1,c_1) \quad \text{and} \quad (a_2,b_2,c_2) \] then the angle \(\theta\) between them is given by: \[ \cos\theta = \frac{a_1a_2+b_1b_2+c_1c_2} {\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}} \] This is the vector dot product formula.

Step 1: Substitute the given direction ratios.
The direction ratios are: \[ (K,3,5) \quad \text{and} \quad (2,-1,2) \] Since the angle between them is \(45^\circ\), \[ \cos45^\circ=\frac{1}{\sqrt2} \] Using the formula: \[ \frac{2K+3(-1)+5(2)} {\sqrt{K^2+3^2+5^2}\sqrt{2^2+(-1)^2+2^2}} = \frac1{\sqrt2} \]

Step 2: Simplify numerator and denominator.
The numerator becomes: \[ 2K-3+10=2K+7 \] The first denominator: \[ \sqrt{K^2+9+25} = \sqrt{K^2+34} \] The second denominator: \[ \sqrt{4+1+4} = \sqrt9 = 3 \] Thus: \[ \frac{2K+7}{3\sqrt{K^2+34}} = \frac1{\sqrt2} \]

Step 3: Cross multiply and solve.
\[ \sqrt2(2K+7)=3\sqrt{K^2+34} \] Squaring both sides: \[ 2(2K+7)^2 = 9(K^2+34) \] Expand: \[ 2(4K^2+28K+49) = 9K^2+306 \] \[ 8K^2+56K+98 = 9K^2+306 \] \[ K^2-56K+208=0 \] Solving: \[ K= \frac{56\pm \sqrt{56^2-4(208)}}{2} \] \[ = \frac{56\pm \sqrt{3136-832}}{2} \] \[ = \frac{56\pm \sqrt{2304}}{2} \] \[ = \frac{56\pm48}{2} \] Thus, \[ K=52 \quad \text{or} \quad K=4 \] Among the given options, only: \[ \boxed{4} \] is present.