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  • Find the equations of the 2 lines through the origin which intersect the line frac{x-3}{2}=frac{y-3}{1}=frac{z}{1} at angles of frac{pi}{3} each.

Find the equations of the 2 lines through the origin which intersect the line \frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1} at angles of \frac{\pi}{3} each.

Answers (1)

Given the equation of the line, we need to find the equations of two lines through the origin which intersect the given line.

According to the theorem, equation of a line with direction ratios d1 = (b1 , b2 , b3 ) that passes through the point (x1 , y1 , z1 ) is  expressed as:

\frac{x-x_{1}}{b_{1}}=\frac{y-y_{1}}{b_{2}}=\frac{z-z_{1}}{b_{3}}

We also know, the angle between two lines with direction ratios d1 and d2 respectively is given by:

\theta = \cos^{-1}\left ( \frac{d_{1}d_{2}}{\left |d_{1} \right |\left |d_{2} \right |} \right )

We use these theorems to find the equations of the two lines.

Let the equation of a line be:

\theta = \cos^{-1}\left ( \frac{d_{1}d_{2}}{\left |d_{1} \right |\left |d_{2} \right |} \right )

Given that it passes through the origin, (0, 0, 0)

Therefore, equation of both lines passing through the origin will be :

\frac{x}{b_{1}}=\frac{y}{b_{2}}=\frac{z}{b_{3}}=\lambda \, \, .....(i)

Let,

\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}=\mu \, \, .....(ii)

Direction ratio of the line = (2, 1, 1)

\Rightarrow d_{1} = (2, 1, 1).... (iii)

If we represent the direction ratio in terms of a position vector,

d_{1}=2\hat{i}+\hat{j}+\hat{k} .....(iv)

Any point on the line is given by (x, y, z). From (ii),

\frac{x-3}{2}=\mu, \frac{y-3}{1}=\mu ,\frac{z}{1}=\mu

\\\text{take} \ \frac{x-3}{2}=\mu\\ \Rightarrow x-3=2\mu\\ \Rightarrow x=2\mu+3\\ \\ take \frac{y-3}{1}=\mu \\ \Rightarrow y-3=\mu\\ \Rightarrow y=\mu+3\\ \\ take \frac{z}{1}=\mu\\ \Rightarrow z=\mu

Hence, any point on line (ii) is P(2\mu + 3, \mu + 3, \mu)

Since line (i) passes through the origin, we can say

\left ( b_{1},b_{2},b_{3} \right )\equiv (2\mu + 3, \mu + 3, \mu)

\Rightarrow \: direction\: \: ratio\: of\; line(i)= (2\mu + 3, \mu + 3, \mu)\\ \Rightarrow d_{2}= (2\mu + 3, \mu + 3, \mu)....(v)

We can represent the direction ratio in terms of position vector like:

d_{2}= \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \: \: ....(vi)

From the theorem, we know

\cos \theta=\frac{d_{1}.d_{2}}{\left |d_{1} \right |\left |d_{2} \right |}

If we substitute the values of d? and d? from (iv) and (vi) in the above equation, and putting \theta=\frac{\pi}{3} from the question:

\Rightarrow \cos \frac{\pi}{3}=\frac{\left ( 2\hat{i}+\hat{j}+\hat{k} \right )\left ( \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right )}{\left | 2\hat{i}+\hat{j}+\hat{k} \right |\left | \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right |}

Solving the numerator,

\left ( 2\hat{i}+\hat{j}+\hat{k} \right )\left ( \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right )= 2\left (2\mu + 3 \right )+1\left ( \mu + 3 \right )+1. \mu

\left ( 2\hat{i}+\hat{j}+\hat{k} \right )\left ( \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right )= 4\mu + 6+\mu + 3+ \mu

\left ( 2\hat{i}+\hat{j}+\hat{k} \right )\left ( \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right )= 4\mu +\mu + \mu+6+3

\left ( 2\hat{i}+\hat{j}+\hat{k} \right )\left ( \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right )= 6\mu+9

Solving the denominator,

\left | 2\hat{i}+\hat{j}+\hat{k} \right |\left | \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right |=\sqrt{2^{2}+1^{2}+1^{2}}\sqrt{ \left (2\mu + 3 \right )^{2}+\left ( \mu + 3 \right )^{2}+ \mu^{2}}

\left | 2\hat{i}+\hat{j}+\hat{k} \right |\left | \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right |=\sqrt{4+1+1}\sqrt{ \left (2\mu \right )^{2}+ 3^{2}+2\left ( 2\mu \right )\left ( 3 \right )+\left ( \mu \right )^{2}+3^{2}+ 2(\mu)(3)+\mu^{2}}

\left | 2\hat{i}+\hat{j}+\hat{k} \right |\left | \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right |=\sqrt{6}\sqrt{4\mu^{2}+9+12\mu+u^{2}+9+6\mu+\mu^{2}}

\left | 2\hat{i}+\hat{j}+\hat{k} \right |\left | \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right |=\sqrt{6}\sqrt{4\mu^{2}+u^{2}+\mu^{2}+12\mu+6\mu+9+9}

\left | 2\hat{i}+\hat{j}+\hat{k} \right |\left | \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right |=\sqrt{6}\sqrt{6\mu^{2}+18\mu+18}

\left | 2\hat{i}+\hat{j}+\hat{k} \right |\left | \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right |=\sqrt{6}\sqrt{6\left (\mu^{2}+3\mu+3 \right )}

\left | 2\hat{i}+\hat{j}+\hat{k} \right |\left | \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right |=\sqrt{6}\sqrt{6}\sqrt{\mu^{2}+3\mu+3}

\left | 2\hat{i}+\hat{j}+\hat{k} \right |\left | \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right |=6\sqrt{\mu^{2}+3\mu+3 }

And cos π/3 = 1/2

Substituting the values, we get

\Rightarrow \frac{1}{2}=\frac{6\mu+9}{6\sqrt{\mu^{2}+3\mu+3}} 

Performing cross multiplication,

\Rightarrow 6\sqrt{\mu^{2}+3\mu+3}=2\left (6\mu+9 \right )\\ \Rightarrow 6\sqrt{\mu^{2}+3\mu+3}=2 \times 3\left (2\mu+3 \right )\\ \Rightarrow 6\sqrt{\mu^{2}+3\mu+3}=6\left (2\mu+3 \right )\\ \Rightarrow \sqrt{\mu^{2}+3\mu+3}=2\mu+3

Squaring both sides,

\Rightarrow \left (\sqrt{\mu^{2}+3\mu+3} \right )^{2}=\left (2\mu+3 \right )^{2}\\ \Rightarrow \mu^{2}+3\mu+3=(2\mu)^{2}+3^{2}+2(2\mu)(3)\left [ \because (a+b)^{2}=a^{2}+b^{2}+2ab \right ]\\ \Rightarrow \mu^{2}+3\mu+3=4\mu^{2}+9+12\mu\\ \Rightarrow 4\mu^{2}-\mu^{2}+12\mu-3\mu+9-3=0\\ \Rightarrow 3\mu^{2}+9\mu+6=0

\Rightarrow 3\left ( \mu^{2}+3\mu+2 \right ) =0\\ \Rightarrow \mu^{2}+3\mu+2=0\\ \Rightarrow\mu^{2}+2\mu+\mu+2=0\\ \Rightarrow \mu\left ( \mu+2 \right )+\left ( \mu+2 \right )=0\\ \Rightarrow \left ( \mu+1 \right )+\left ( \mu+2 \right )=0\\

\Rightarrow \left ( \mu+1 \right )=0 \: \: or\: \left ( \mu+2 \right )=0\\ \Rightarrow \mu=-1 \: \: \: or\: \: \: \mu=-2

Therefore, from equation (v)

Direction ratio =(2\mu+ 3, \mu + 3, \mu)

Putting μ = -1:

Direction Ratio = (2(-1) + 3, (-1) + 3, -1)

⇒ Direction Ratio = (-2 + 3, -1 + 3, -1)

⇒ Direction Ratio = (1, 2, -1) …(vi)

Now putting μ = -2:

Direction Ratio = (2(-2) + 3, (-2) + 3, -2)

⇒ Direction Ratio = (-4 + 3, -2 + 3, -2)

⇒ Direction Ratio = (-1, 1, -2) …(vii)

Using the direction ratios in (vi) and (vii) in equation (i);

\frac{x}{b_{1}}=\frac{y}{b_{2}}=\frac{z}{b_{3}}=\lambda\\ \\ \\ \frac{x}{1}=\frac{y}{2}=\frac{z}{-1}=\lambda

And,

\frac{x}{-1}=\frac{y}{1}=\frac{z}{-2}=\lambda

Therefore, the two required lines are \frac{x}{1}=\frac{y}{2}=\frac{z}{-1}=\lambda and \frac{x}{-1}=\frac{y}{1}=\frac{z}{-2}=\lambda

 

 

Posted by

infoexpert24

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