Question

Find the angle between the lines whose direction cosines are given by the equations.
(i) l + 2m +3n = 0 and 3lm – 4ln + mn = 0
(ii) l + m + n = 0 and l2 + m2 – n2 = 0

Answer 1

Answer on Question #46181 – Math - Vector Calculus

Problem.

Find the angle between the lines whose direction cosines are given by the equations.

(i) $l + 2m + 3n = 0$ and $3lm - 4ln + mn = 0$

(ii) $l + m + n = 0$ and $l^2 + m^2 - n^2 = 0$

Remark: I suppose that the statement is incorrectly formatted. The correct statement is:

"Find the angle between the lines whose direction cosines are given by the equations.

(i) $l + 2m + 3n = 0$ and $3lm - 4ln + mn = 0$

(ii) $l + m + n = 0$ and $l^2 + m^2 - n^2 = 0$"

Solution:

(i) If $l, m, n$ are direction cosines, then $l^2 + m^2 + n^2 = 1$.

$$\begin{array}{l} l = -2m - 3n, \text{ so } 3(-2m - 3n)m - 4(-2m - 3n)n + mn = 0. \text{ Hence} \\ \quad -6m^2 - 9mn + 8mn + 12n^2 + mn = 0 \text{ and } 12n^2 = 6m^2 \text{ or } 2n^2 = m^2. \text{ Then } m = \\ \quad \sqrt{2}n \text{ and } m = -\sqrt{2}n. \end{array}$$ $$l^2 + m^2 + n^2 = 1 \text{ and } l = -2m - 3n, \text{ so } 4m^2 + 12mn + 9n^2 + m^2 + n^2 = 1. \text{ Hence} \quad 5m^2 + 12mn + 10n^2 = 20n^2 + 12mn = 1.$$ $$\text{If } m = \sqrt{2}n, \text{ then } 20n^2 + 12\sqrt{2}n^2 = 1 \text{ or } n^2(20 + 12\sqrt{2}) = 1. \text{ Then } n = \frac{1}{\sqrt{20 + 12\sqrt{2}}} \text{ or } n = -\frac{1}{\sqrt{20 + 12\sqrt{2}}}. \text{ We obtain two triples } (l, m, n) \text{ of direction cosines}$$ $$\left(\frac{-2\sqrt{2} - 3}{\sqrt{20 + 12\sqrt{2}}}, \frac{\sqrt{2}}{\sqrt{20 + 12\sqrt{2}}}, \frac{1}{\sqrt{20 + 12\sqrt{2}}}\right), \left(\frac{2\sqrt{2} + 3}{\sqrt{20 + 12\sqrt{2}}}, -\frac{\sqrt{2}}{\sqrt{20 + 12\sqrt{2}}}, -\frac{1}{\sqrt{20 + 12\sqrt{2}}}\right).$$

If $m = -\sqrt{2}n$, then $20n^2 - 12\sqrt{2}n^2 = 1$ or $n^2(20 - 12\sqrt{2}) = 1$. Then $n = \frac{1}{\sqrt{20 - 12\sqrt{2}}}$ or $n = -\frac{1}{\sqrt{20 - 12\sqrt{2}}}$. We obtain two triples $(l, m, n)$ of direction cosines $\left(\frac{2\sqrt{2} - 3}{\sqrt{20 - 12\sqrt{2}}}, -\frac{\sqrt{2}}{\sqrt{20 - 12\sqrt{2}}}, \frac{1}{\sqrt{20 - 12\sqrt{2}}}\right)$, $\left(\frac{-2\sqrt{2} + 3}{\sqrt{20 - 12\sqrt{2}}}, \frac{\sqrt{2}}{\sqrt{20 - 12\sqrt{2}}}, -\frac{1}{\sqrt{20 - 12\sqrt{2}}}\right)$.

Answer: $\left(\frac{-2\sqrt{2} - 3}{\sqrt{20 + 12\sqrt{2}}}, \frac{\sqrt{2}}{\sqrt{20 + 12\sqrt{2}}}, \frac{1}{\sqrt{20 + 12\sqrt{2}}}\right), \left(\frac{2\sqrt{2} + 3}{\sqrt{20 + 12\sqrt{2}}}, -\frac{\sqrt{2}}{\sqrt{20 + 12\sqrt{2}}}, -\frac{1}{\sqrt{20 + 12\sqrt{2}}}\right)$, $\left(\frac{2\sqrt{2} - 3}{\sqrt{20 - 12\sqrt{2}}}, -\frac{\sqrt{2}}{\sqrt{20 - 12\sqrt{2}}}, \frac{1}{\sqrt{20 - 12\sqrt{2}}}\right), \left(\frac{-2\sqrt{2} + 3}{\sqrt{20 - 12\sqrt{2}}}, \frac{\sqrt{2}}{\sqrt{20 - 12\sqrt{2}}}, -\frac{1}{\sqrt{20 - 12\sqrt{2}}}\right)$.

(ii) If $l, m, n$ are direction cosines, then $l^2 + m^2 + n^2 = 1$. If $l^2 + m^2 = n^2$ and $l^2 + m^2 + n^2 = 1$, then $2n^2 = 1$ and $l^2 + m^2 = \frac{1}{2}$. $l + m = -n$, so $l^2 + 2lm + m^2 = n^2$. Hence $2lm = 0$ ($l = 0$ or $m = 0$) and $n^2 = \frac{1}{2}$.

If $l = 0$, then $m = \frac{1}{\sqrt{2}}$ and $n = -\frac{1}{\sqrt{2}}$ or $m = -\frac{1}{\sqrt{2}}$ and $n = \frac{1}{\sqrt{2}}$.

If $m = 0$, then $n = \frac{1}{\sqrt{2}}$ and $l = -\frac{1}{\sqrt{2}}$ or $l = -\frac{1}{\sqrt{2}}$ and $n = \frac{1}{\sqrt{2}}$.

Therefore, we obtain three four triples $(l, m, n)$ of direction cosines $\left(0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right), \left(0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$, $\left(-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right), \left(\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}}\right)$.

Answer: $\left(0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right), \left(0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right), \left(-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right), \left(\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}}\right)$.

www.AssignmentExpert.com