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  • Find the angle between the lines whose direction cosines are given by equations l + m + n = 0, l² + m² - n² = 0.

Find the angle between the lines whose direction cosines are given by equations l + m + n = 0, l² + m² - n² = 0.

 

 

Answers (1)

Given, two lines whose direction cosines are l + m + n = 0 - (i); and l² + m² - n² = 0 - (ii). We need to find the angle between these lines.

First, we must find the values of l, m and n. 

From equation (i), l + m + n = 0

=> n = - l - m

=> n = -(l + m) …(iii)

If we substitute the value of n from (i) in (ii),
 l^{2}+m^{2}-n^{2}=0\\ \Rightarrow l^{2}+m^{2}-\left (-\left ( l+m \right ) \right )^{2}=0\\ \Rightarrow l^{2}+m^{2}-\left ( l+m \right ) ^{2}=0\\ \Rightarrow l^{2}+m^{2}-\left ( l^{2}+m^{2}+2lm \right )=0\\ \Rightarrow l^{2}+m^{2} -l^{2}-m^{2}-2lm=0\\ \Rightarrow l^{2}-l^{2}+m^{2}-m^{2}-2lm=0\\ \Rightarrow -2lm=0\\ \Rightarrow lm=0

⇒ l = 0 or m = 0

Putting l = 0 in equation (i),

=> 0 + m + n = 0

=> m + n = 0

=> m = -n

If m = \lambda, then

n = -m = -\lambda

Hence, direction ratios (l, m, n) = (0, \lambda, -\lambda)

=> Position vector parallel to these given lines = 0\hat{i}+\lambda\hat{j}-\lambda \hat{k}

\Rightarrow d_{1}=\lambda\hat{j}-\lambda \hat{k}

Now, putting m = 0 in equation (i),

=> l + 0 + n = 0

=> l + n = 0

=> l = -n

If n = \lambda, then

l = -n = -\lambda

Hence, direction ratios (l, m, n) = (-\lambda, 0, \lambda)

=> Position vector parallel to these given lines = -\lambda \hat{i}+\0\hat{j}+\lambda \hat{k}

\Rightarrow d_{2}=-\lambda\hat{i}+\lambda \hat{k}

From the theorem, we get the angle between the two lines whose direction ratios are d1 and d2 as:

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

If we substitute the values of d1 and d2, we get

\theta=\cos^{-1}\left ( \frac{\left| \left ( \lambda \hat{j}-\lambda \hat{k} \right )\left (- \lambda \hat{i}+\lambda \hat{k} \right ) \right |}{\left | \left ( \lambda \hat{j}-\lambda \hat{k} \right )\right |\left |\left (- \lambda \hat{i}+\lambda \hat{k} \right ) \right |} \right )

Solving the numerator,

\left ( \lambda \hat{j}-\lambda \hat{k} \right )\left (- \lambda \hat{i}+\lambda \hat{k} \right ) =0+0+\left ( -\lambda \right )\left ( \lambda \right )\\ \Rightarrow \left ( \lambda \hat{j}-\lambda \hat{k} \right )\left (- \lambda \hat{i}+\lambda \hat{k} \right )= -\lambda^{2}

Solving the denominator,

\left |\left ( \lambda \hat{j}-\lambda \hat{k} \right ) \right |\left |\left (- \lambda \hat{i}+\lambda \hat{k} \right ) \right |=\sqrt{ \lambda ^{2}\left (-\lambda \right )^{2}} \sqrt{\left (- \lambda \right )^{2}+\lambda ^{2}}

\left |\left ( \lambda \hat{j}-\lambda \hat{k} \right ) \right |\left |\left (- \lambda \hat{i}+\lambda \hat{k} \right ) \right |=\sqrt{ \lambda ^{2}+\lambda^{2}} \sqrt{ \lambda^{2}+\lambda ^{2}}

\left |\left ( \lambda \hat{j}-\lambda \hat{k} \right ) \right |\left |\left (- \lambda \hat{i}+\lambda \hat{k} \right ) \right |= \lambda ^{2}+\lambda^{2}

\left |\left ( \lambda \hat{j}-\lambda \hat{k} \right ) \right |\left |\left (- \lambda \hat{i}+\lambda \hat{k} \right ) \right |=2 \lambda ^{2}

Substituting the values in θ,

\theta=\cos^{-1}\left ( \frac{\left | -\lambda^{2} \right |}{\lambda^{2}} \right )\\ \Rightarrow \theta=\cos^{-1}=\frac{1}{2}\\ \Rightarrow \theta=\frac{\pi}{3}\left [ \because \cos\frac{\pi}{3}=\frac{1}{2} \right ]

Therefore, the required angle between the lines is π/3.

Posted by

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