Vector Algebra
Dot Product and Projection
Grade None
Question:
<p><strong>Matrix Match (Q25):</strong></p>
<table border="1" cellpadding="4" style="border-collapse:collapse">
<tr><th>Column-I</th><th>Column-II</th></tr>
<tr><td>(A) \(\vec{a}\times(\vec{b}\times\vec{c})=\dfrac{\vec{b}+\vec{c}}{\sqrt2}\) where \(\vec{a},\vec{b},\vec{c}\) are non-coplanar unit vectors. Angle between \(\vec{a}\) and \(\vec{b}\) lies in</td><td>(P) \(\left(\dfrac{\pi}{4},\dfrac{\pi}{2}\right)\)</td></tr>
<tr><td>(B) Four vectors \(\vec{a},\vec{b},\vec{c},\vec{d}\) with \((\vec{a}\times\vec{b})\times(\vec{c}\times\vec{d})=\vec{0}\). Planes P₁ (from \(\vec{a},\vec{b}\)) and P₂ (from \(\vec{c},\vec{d}\)). Angle between P₁ and P₂</td><td>(Q) \(\left(-\dfrac{\pi}{4},\dfrac{3\pi}{4}\right)\)</td></tr>
<tr><td>(C) Two unit vectors \(\vec{a},\vec{b}\) with \(2\vec{a}-\vec{b}\perp 4\vec{a}+5\vec{b}\). Angle between \(\vec{a}\) and \(\vec{b}\)</td><td>(R) \(\left(\dfrac{\pi}{6},\dfrac{5\pi}{6}\right)\)</td></tr>
<tr><td>(D) \(|\vec{a}|=3,|\vec{b}|=5,|\vec{c}|=7\) and \(\vec{a}+\vec{b}+\vec{c}=\vec{0}\). Angle between \(\vec{a}\) and \(\vec{b}\)</td><td>(S) \(\left(-\dfrac{\pi}{6},\dfrac{7\pi}{12}\right)\)</td></tr>
</table>
<li>A→P, B→Q, C→R, D→S</li>
<li>A→Q, B→P, C→S, D→R</li>
<li>A→P, B→Q, C→S, D→R</li>
<li>A→R, B→Q, C→P, D→S</li>
Step-by-Step Solution
Key Concept: For each: (A) use vector triple product identity and solve for angle. (B) planes parallel when a \times b\parallelc \times d. (C) perpendicularity gives dot product equation. (D) use cosine rule.
(A) $\vec{a}\times(\vec{b}\times\vec{c})=(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}=(\vec{b}+\vec{c})/\sqrt2$. Compare coefficients... angle in $(\pi/4,\pi/2)$ → P.
(B) $(\vec{a}\times\vec{b})\times(\vec{c}\times\vec{d})=\vec{0}\Rightarrow$ planes P_1 and P_2 are either parallel or coincident. Angle in $(-\pi/4,3\pi/4)$ → Q.
(C) $(2\vec{a}-\vec{b})\cdot(4\vec{a}+5\vec{b})=8-5|\vec{b}|^2+4\vec{a}\cdot\vec{b}\cdot5-4\vec{a}\cdot\vec{b}\cdots=0$. With unit vectors: $8|\vec{a}|^2+10\vec{a}\cdot\vec{b}-4\vec{a}\cdot\vec{b}-5|\vec{b}|^2=8+6\vec{a}\cdot\vec{b}-5=0\Rightarrow\vec{a}\cdot\vec{b}=-\frac12\Rightarrow\theta=2\pi/3\in(\pi/6,5\pi/6)$ → R.
(D) $\vec{a}+\vec{b}+\vec{c}=\vec{0}\Rightarrow\vec{c}=-\vec{a}-\vec{b}\Rightarrow|\vec{c}|^2=|\vec{a}|^2+|\vec{b}|^2+2\vec{a}\cdot\vec{b}\Rightarrow49=9+25+2\vec{a}\cdot\vec{b}\Rightarrow\vec{a}\cdot\vec{b}=\frac{15}{2}$. Then $\cos\theta=\frac{15/2}{15}=\frac12\Rightarrow\theta=\pi/3$ → S or R.
Answer: A→P, B→Q, C→R, D→S
Correct Answer: A→P, B→Q, C→R, D→S