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# Show that the three lines with direction cosines

Question:

Show that the three lines with direction cosines

$\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13} ; \frac{4}{13}, \frac{12}{13}, \frac{3}{13} ; \frac{3}{13}, \frac{-4}{13}, \frac{12}{13}$ are mutually perpendicular.

Solution:

Two lines with direction cosines, l1m1n1 and l2m2n2, are perpendicular to each other, if l1l2 + m1m2 + n1n2 = 0

(i) For the lines with direction cosines, $\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}$ and $\frac{4}{13}, \frac{12}{13}, \frac{3}{13}$, we obtain

$l_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}=\frac{12}{13} \times \frac{4}{13}+\left(\frac{-3}{13}\right) \times \frac{12}{13}+\left(\frac{-4}{13}\right) \times \frac{3}{13}$

$=\frac{48}{169}-\frac{36}{169}-\frac{12}{169}$

$=0$

Therefore, the lines are perpendicular.

(ii) For the lines with direction cosines, $\frac{4}{13}, \frac{12}{13}, \frac{3}{13}$ and $\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}$, we obtain

$l_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}=\frac{4}{13} \times \frac{3}{13}+\frac{12}{13} \times\left(\frac{-4}{13}\right)+\frac{3}{13} \times \frac{12}{13}$

$=\frac{12}{169}-\frac{48}{169}+\frac{36}{169}$

$=0$

Therefore, the lines are perpendicular.

(iii) For the lines with direction cosines, $\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}$ and $\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}$, we obtain

$l_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}=\left(\frac{3}{13}\right) \times\left(\frac{12}{13}\right)+\left(\frac{-4}{13}\right) \times\left(\frac{-3}{13}\right)+\left(\frac{12}{13}\right) \times\left(\frac{-4}{13}\right)$

$=\frac{36}{169}+\frac{12}{169}-\frac{48}{169}$

$=0$

Therefore, the lines are perpendicular.

Thus, all the lines are mutually perpendicular.