Question

Find the square of the vector \[ (4\mathbf{i} + 3\mathbf{j})^2. \]

• A. 7
• B. 19
• C. 25
• D. 49

Hint:

When squaring a vector, use the dot product. The square of the vector is the sum of the squares of its components, considering the dot products of orthogonal vectors are zero.

Answer 1

The Correct Option is C

The square of the vector \( \mathbf{A} = 4 \mathbf{i} + 3 \mathbf{j} \) is calculated as: \[ |\mathbf{A}|^2 = (4\mathbf{i} + 3\mathbf{j}) \cdot (4\mathbf{i} + 3\mathbf{j}). \] Using the distributive property of the dot product: \[ |\mathbf{A}|^2 = 4^2(\mathbf{i} \cdot \mathbf{i}) + 4 \times 3 (\mathbf{i} \cdot \mathbf{j}) + 3^2 (\mathbf{j} \cdot \mathbf{j}). \] Since \( \mathbf{i} \cdot \mathbf{i} = 1 \), \( \mathbf{j} \cdot \mathbf{j} = 1 \), and \( \mathbf{i} \cdot \mathbf{j} = 0 \) (since the vectors are orthogonal): \[ |\mathbf{A}|^2 = 16 \times 1 + 0 + 9 \times 1 = 16 + 9 = 25. \] Thus, the correct answer is option (C) 25.