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Question
Show that the straight lines whose direction cosines are given by 2l + 2m – n = 0 and mn + nl + lm = 0 are at right angles.
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Solution
Given that 2l + 2m – n = 0 ......(i)
And mn + nl + lm = 0 .....(ii)
Eliminating m from equation (i) and (ii) we get,
m = `("n" - 2l)/2` .....[From (i)]
⇒ `(("n" - 2l)/2)"n" + "nl" + "l"(("n" - 2"l")/2)` = 0
⇒ `("n"^2 - 2"n"l + 2"n"l + "nl" - 2"l"^2)/2` = 0
⇒ n2 + nl – 2l 2 = 0
⇒ n2 + 2nl – nl – 2l2 = 0
⇒ n(n + 2l) – l(n + 2l) = 0
⇒ (n – l)(n + 2l) = 0
⇒ n = – 2l and n = l
∴ m = `(-2l - 2l)/2`, m = `(l - 2l)/2`
⇒ m = –2l, m = `(-1)/2`
Therefore, the direction ratios are proportional to l, – 2l, –2l and l, `(-1)/2`, l.
⇒ 1, – 2, – 2 and 2, – 1, 2
If the two lines are perpendicular to each other then
1(2) – 2(– 1) – 2 × 2 = 0
2 + 2 – 4 = 0
0 = 0
Hence, the two lines are perpendicular.
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