Question 67 of 68intermediate🔧 ApplyMCQ1 mark

The value of î ⋅ (ĵ × k̂) + ĵ ⋅ (î × k̂) + k̂ ⋅ (î × ĵ) is

Exercise: MISCELLANEOUS EXERCISE ON CHAPTER 10 | Q: 18 | (Page 36)
For More Understanding

Explanation

The correct answer is C. Using the cross product identities from the textbook: i^×j^=k^\hat{i} \times \hat{j} = \hat{k}, j^×k^=i^\hat{j} \times \hat{k} = \hat{i}, k^×i^=j^\hat{k} \times \hat{i} = \hat{j}, and i^×k^=−j^\hat{i} \times \hat{k} = -\hat{j}. The dot product of unit vectors yields 1 for same vectors. Thus each term evaluates to 1, -1, 1 respectively, summing to 1.

Solution Steps

  1. Step 1: Compute j^×k^=i^\hat{j} \times \hat{k} = \hat{i} (from context). Then i^⋅(j^×k^)=i^⋅i^=1\hat{i} \cdot (\hat{j} \times \hat{k}) = \hat{i} \cdot \hat{i} = 1.

  2. Step 2: Compute i^×k^=−j^\hat{i} \times \hat{k} = -\hat{j} (since i^×k^=−k^×i^\hat{i} \times \hat{k} = -\hat{k} \times \hat{i} and k^×i^=j^\hat{k} \times \hat{i} = \hat{j}). Then j^⋅(i^×k^)=j^⋅(−j^)=−1\hat{j} \cdot (\hat{i} \times \hat{k}) = \hat{j} \cdot (-\hat{j}) = -1.

  3. Step 3: Compute i^×j^=k^\hat{i} \times \hat{j} = \hat{k} (from context). Then k^⋅(i^×j^)=k^⋅k^=1\hat{k} \cdot (\hat{i} \times \hat{j}) = \hat{k} \cdot \hat{k} = 1.

  4. Step 4: Sum the three results: 1+(−1)+1=11 + (-1) + 1 = 1. Hence the value is 1, corresponding to option (C).

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