Vector Algebra
Dot Product and Projection
Grade 12
Question:
<p>Which of the following statements hold good for vectors \(\vec{a},\vec{b}\)?</p>
<li>Triangle inequality: \(|\vec{a}+\vec{b}|\leq|\vec{a}|+|\vec{b}|\)</li>
<li>Cauchy-Schwarz: \(|\vec{a}\cdot\vec{b}|\leq|\vec{a}||\vec{b}|\)</li>
<li>Lagrange: \(|\vec{a}\times\vec{b}|^2+(\vec{a}\cdot\vec{b})^2=|\vec{a}|^2|\vec{b}|^2\)</li>
<li>\((\vec{a}+\vec{b})\times(\vec{a}-\vec{b})=2(\vec{b}\times\vec{a})\)</li>
Step-by-Step Solution
Key Concept: All four are standard vector identities. Verify each: triangle inequality, Cauchy-Schwarz, Lagrange identity, and cross product expansion.
(A) Triangle inequality: standard — always true. ✓
(B) Cauchy-Schwarz: \(|\vec{a}\cdot\vec{b}|=|\vec{a}||\vec{b}||\cos\theta|\leq|\vec{a}||\vec{b}|\). ✓
(C) Lagrange: \(|\vec{a}\times\vec{b}|^2=|\vec{a}|^2|\vec{b}|^2\sin^2\theta\), \((\vec{a}\cdot\vec{b})^2=|\vec{a}|^2|\vec{b}|^2\cos^2\theta\). Sum \(=|\vec{a}|^2|\vec{b}|^2\). ✓
(D) \((\vec{a}+\vec{b})\times(\vec{a}-\vec{b})=\vec{a}\times\vec{a}-\vec{a}\times\vec{b}+\vec{b}\times\vec{a}-\vec{b}\times\vec{b}=0-\vec{a}\times\vec{b}-\vec{a}\times\vec{b}-0=-2\vec{a}\times\vec{b}=2\vec{b}\times\vec{a}\). ✓
Answer: ABCD
Correct Answer: ABCD