Chapter 25: Problem 65
Let \(\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}, \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\) and \(\overrightarrow{\mathrm{c}}\) be a vector such that \(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{b}}=\overrightarrow{0}\) and \(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=4\), then \(|\overrightarrow{\mathrm{c}}|^{2}\) is equal to: [Jan 09, 2019] (a) \(\frac{19}{2}\) (b) 9 (c) 8 (d) \(\frac{17}{2}\)
Short Answer
Step by step solution
Express the Given Equation
Analyze the Cross Product Equation
Solve the System of Equations
Find Values for x, y, and z
Calculate \(|\overrightarrow{\mathrm{c}}|^2\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cross Product
To calculate the cross product, use the components of the vectors:
- \( \overrightarrow{\mathbf{u}} = u_1 \hat{i} + u_2 \hat{j} + u_3 \hat{k} \)
- \( \overrightarrow{\mathbf{v}} = v_1 \hat{i} + v_2 \hat{j} + v_3 \hat{k} \)
Dot Product
This operation results in a scalar rather than a vector. The formula is as follows:
- If \( \overrightarrow{\mathbf{u}} = u_1 \hat{i} + u_2 \hat{j} + u_3 \hat{k} \)
- And \( \overrightarrow{\mathbf{v}} = v_1 \hat{i} + v_2 \hat{j} + v_3 \hat{k} \)
Vector Magnitude
- The magnitude is never negative, as it reflects length.
- It's a real number that offers insight into how vectors compare in size.