Find the direction cosines of the vector î + 2ĵ + 3k̂.

Question 15 of 68beginner🔧 ApplyNumerical1 mark

Correct Answer

The direction cosines of the vector î + 2ĵ + 3k̂ are (1/ $\sqrt{14}$ , 2/ $\sqrt{14}$ , 3/ $\sqrt{14}$ ).

Exercise: EXERCISE 10.2 | Q: 12 | (Page 17)

For More Understanding

Explanation

This question tests the understanding of direction cosines as explained in the textbook context. The direction cosines of a vector are the cosines of the angles that the vector makes with the positive x, y, and z-axes. According to Example 9 in the context, the direction ratios (a, b, c) of a vector r = xî + yĵ + zk̂ are simply its components (x, y, z). The direction cosines (l, m, n) are then calculated by dividing each direction ratio by the magnitude of the vector.

Solution Steps

Step 1: Identify the direction ratios of the vector î + 2ĵ + 3k̂. From the context, direction ratios are the respective components of the vector. So, a = 1, b = 2, c = 3.
Step 2: Calculate the magnitude of the vector. |r⃗| = $\sqrt{1² + 2² + 3²}$ = $\sqrt{1 + 4 + 9}$ = $\sqrt{14}$ .
Step 3: Apply the formula for direction cosines from the context. If l, m, n are direction cosines, then l = a/|r⃗|, m = b/|r⃗|, n = c/|r⃗|.
Step 4: Calculate each direction cosine. l = 1/ $\sqrt{14}$ , m = 2/ $\sqrt{14}$ , n = 3/ $\sqrt{14}$ .
Step 5: Write the direction cosines as (1/ $\sqrt{14}$ , 2/ $\sqrt{14}$ , 3/ $\sqrt{14}$ ).

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