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The lines having directional derivatives as a1a2+b1b2+c1c2=0 |
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Answer» Answer: 1. SHOW that the three lines with direction cosines Are mutually perpendicular. Solution: Let us consider the direction cosines of L1, L2 and L3 be l1, m1, N1; l2, m2, n2 and l3, M3, n3. We know that If l1, m1, n1 and l2, m2, n2 are the direction cosines of two lines; And θ is the acute angle between the two lines; Then COS θ = |l1l2 + m1m2 + n1n2| If two lines are perpendicular, then the angle between the two is θ = 90° For perpendicular lines, | l1l2 + m1m2 + n1n2 | = cos 90° = 0, i.e. | l1l2 + m1m2 + n1n2 | = 0 So, in order to check if the three lines are mutually perpendicular, we compute | l1l2 + m1m2 + n1n2 | for all the pairs of the three lines. Firstly let us compute, | l1l2 + m1m2 + n1n2 | So, L1⊥ L2 …… (1) Similarly, Let us compute, | l2l3 + m2 |
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