arthham

JEE Advanced 2015 · Vector Algebra

Let ΔPQR\Delta PQR be a triangle. Let a⃗=QR→\vec{a} = \overrightarrow{QR}, b⃗=RP→\vec{b} = \overrightarrow{RP} and c⃗=PQ→\vec{c} = \overrightarrow{PQ}. If ∣a⃗∣=12|\vec{a}| = 12, ∣b⃗∣=43|\vec{b}| = 4\sqrt{3}, b⃗⋅c⃗=24\vec{b} \cdot \vec{c} = 24, then which of the following is (are) true?

Answers: A. ∣c⃗∣22−∣a⃗∣=12\frac{|\vec{c}|^{2}}{2} - |\vec{a}| = 12 · C. ∣a⃗×b⃗+c⃗×a⃗∣=483|\vec{a} \times \vec{b} + \vec{c} \times \vec{a}| = 48\sqrt{3} · D. a⃗⋅b⃗=−72\vec{a} \cdot \vec{b} = -72

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ΔPQR: a⃗=QR→, b⃗=RP→, c⃗=PQ→\Delta PQR:\ \vec{a} = \overrightarrow{QR},\ \vec{b} = \overrightarrow{RP},\ \vec{c} = \overrightarrow{PQ} ∣a⃗∣=12,∣b⃗∣=43,b⃗⋅c⃗=24|\vec{a}| = 12,\quad |\vec{b}| = 4\sqrt{3},\quad \vec{b} \cdot \vec{c} = 24 To find: which of (1),(2),(3),(4) are true\text{To find: which of }(1),(2),(3),(4)\text{ are true} a⃗=QR→,b⃗=RP→,c⃗=PQ→\vec{a} = \overrightarrow{QR}, \quad \vec{b} = \overrightarrow{RP}, \quad \vec{c} = \overrightarrow{PQ} a⃗+b⃗+c⃗=0⃗\vec{a} + \vec{b} + \vec{c} = \vec{0} Plan: 1. a⃗+b⃗+c⃗=0⃗⇒a⃗⋅b⃗\text{Plan: } 1.\ \vec{a} + \vec{b} + \vec{c} = \vec{0} \Rightarrow \vec{a} \cdot \vec{b} 2. ∣c⃗∣2=∣a⃗+b⃗∣22.\ |\vec{c}|^{2} = |\vec{a} + \vec{b}|^{2} 3. c⃗×a⃗=a⃗×b⃗3.\ \vec{c} \times \vec{a} = \vec{a} \times \vec{b} ⇒a⃗⋅b⃗+b⃗⋅b⃗+b⃗⋅c⃗=0\Rightarrow \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} + \vec{b} \cdot \vec{c} = 0 a⃗⋅b⃗+48+24=0\vec{a} \cdot \vec{b} + 48 + 24 = 0 a⃗⋅b⃗=−72⇒ option (4) is true\vec{a} \cdot \vec{b} = -72 \quad \Rightarrow \ \text{option (4) is true} c⃗=−(a⃗+b⃗)⇒∣c⃗∣2=∣a⃗+b⃗∣2\vec{c} = -(\vec{a} + \vec{b}) \Rightarrow |\vec{c}|^{2} = |\vec{a} + \vec{b}|^{2} =∣a⃗∣2+∣b⃗∣2+2a⃗⋅b⃗= |\vec{a}|^{2} + |\vec{b}|^{2} + 2\vec{a} \cdot \vec{b} ∣c⃗∣2=144+48−144=48|\vec{c}|^{2} = 144 + 48 - 144 = 48 ∣c⃗∣22−∣a⃗∣=482−12\dfrac{|\vec{c}|^{2}}{2} - |\vec{a}| = \dfrac{48}{2} - 12 =24−12=12✓= 24 - 12 = 12 \quad \checkmark ∣c⃗∣22+∣a⃗∣=24+12=36\dfrac{|\vec{c}|^{2}}{2} + |\vec{a}| = 24 + 12 = 36 36≠30so option (2) is false36 \neq 30 \quad \text{so option (2) is false} c⃗×a⃗=−(a⃗+b⃗)×a⃗\vec{c} \times \vec{a} = -(\vec{a} + \vec{b}) \times \vec{a} =−(a⃗×a⃗+b⃗×a⃗)=a⃗×b⃗= -(\vec{a} \times \vec{a} + \vec{b} \times \vec{a}) = \vec{a} \times \vec{b} a⃗×b⃗+c⃗×a⃗=2(a⃗×b⃗)\vec{a} \times \vec{b} + \vec{c} \times \vec{a} = 2(\vec{a} \times \vec{b}) ∣a⃗×b⃗+c⃗×a⃗∣=2∣a⃗×b⃗∣|\vec{a} \times \vec{b} + \vec{c} \times \vec{a}| = 2|\vec{a} \times \vec{b}| ∣a⃗×b⃗∣2=∣a⃗∣2∣b⃗∣2−(a⃗⋅b⃗)2|\vec{a} \times \vec{b}|^{2} = |\vec{a}|^{2}|\vec{b}|^{2} - (\vec{a} \cdot \vec{b})^{2} =144⋅48−(−72)2=1728= 144 \cdot 48 - (-72)^{2} = 1728 ∣a⃗×b⃗∣=1728=243|\vec{a} \times \vec{b}| = \sqrt{1728} = 24\sqrt{3} ∣a⃗×b⃗+c⃗×a⃗∣=2⋅243=483✓|\vec{a} \times \vec{b} + \vec{c} \times \vec{a}| = 2 \cdot 24\sqrt{3} = 48\sqrt{3} \quad \checkmark True options: (1), (3), (4)\text{True options: } (1),\ (3),\ (4)

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