Question 16 of 68intermediate🔧 ApplyNumerical2 marks

Find the direction cosines of the vector joining the points A (1, 2, -3) and B (-1, -2, 1), directed from A to B.

Correct Answer

The direction cosines of the vector directed from A to B are (−13,−23,23)\left(-\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}\right).

Exercise: EXERCISE 10.2 | Q: 13 | (Page 17)
For More Understanding

Explanation

This question tests the concept of finding direction cosines of a vector joining two points. From the textbook context, the vector joining points P₁(x₁,y₁,z₁) and P₂(x₂,y₂,z₂) is given by (x₂-x₁)i + (y₂-y₁)j + (z₂-z₁)k. The direction cosines are obtained by dividing each component by the magnitude, as stated: cos α = a₁/|a|, cos β = a₂/|a|, cos γ = a₃/|a|.

Solution Steps

  1. Step 1: Find the vector AB = (-1-1)i + (-2-2)j + (1-(-3))k = -2i - 4j + 4k

  2. Step 2: Calculate the magnitude |AB| = (−2\sqrt{(-2}² + (-4)² + (4)²) = 4+16+16\sqrt{4+16+16} = 36\sqrt{36} = 6

  3. Step 3: Find direction cosines by dividing each component by magnitude: (-2/6, -4/6, 4/6) = (-1/3, -2/3, 2/3)

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