arthham

JEE Advanced 2012 · Vector Algebra

If aa and bb are vectors such that ∣a+b∣=29|a+b| = \sqrt{29} and a×(2i+3j+4k)=(2i+3j+4k)×ba \times (2i+3j+4k) = (2i+3j+4k) \times b, then a possible value of (a+b)⋅(−7i+2j+3k)(a+b) \cdot (-7i+2j+3k) is

Answer: C. 44

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Written solution

If a⃗ and b⃗ are vectors such that |a⃗+b⃗|=√29 and a⃗×(2î+3ĵ+4k̂)=(2î+3ĵ+4k̂)×b⃗, then a possible value of (a⃗+b⃗)·(-7î+2ĵ+3k̂) is

∣a⃗+b⃗∣=29\left|\vec a+\vec b\right| = \sqrt{29}

a⃗×c⃗=c⃗×b⃗,c⃗=2i^+3j^+4k^\vec a\times\vec c = \vec c\times\vec b,\quad \vec c = 2\hat i+3\hat j+4\hat k

Find a possible value of (a⃗+b⃗)⋅(−7i^+2j^+3k^)\text{Find a possible value of } (\vec a+\vec b)\cdot(-7\hat i+2\hat j+3\hat k)

1. (a⃗+b⃗)×c⃗=0⃗⇒(a⃗+b⃗)∥c⃗1.\ (\vec a+\vec b)\times\vec c = \vec 0 \Rightarrow (\vec a+\vec b)\parallel\vec c

2. ∣a⃗+b⃗∣=∣c⃗∣⇒a⃗+b⃗=±c⃗2.\ |\vec a+\vec b|=|\vec c| \Rightarrow \vec a+\vec b=\pm\vec c

3. evaluate c⃗⋅(−7i^+2j^+3k^)3.\ \text{evaluate } \vec c\cdot(-7\hat i+2\hat j+3\hat k)

a⃗×c⃗=c⃗×b⃗=−b⃗×c⃗\vec a\times\vec c = \vec c\times\vec b = -\vec b\times\vec c

⇒ a⃗×c⃗+b⃗×c⃗=(a⃗+b⃗)×c⃗=0⃗\Rightarrow\ \vec a\times\vec c + \vec b\times\vec c = (\vec a+\vec b)\times\vec c = \vec 0

a⃗+b⃗∥c⃗\vec a+\vec b \parallel \vec c

∣c⃗∣=4+9+16=29=∣a⃗+b⃗∣⇒a⃗+b⃗=±c⃗|\vec c| = \sqrt{4+9+16} = \sqrt{29} = |\vec a+\vec b| \Rightarrow \vec a+\vec b = \pm\vec c

c⃗⋅w⃗=2(−7)+3(2)+4(3)=−14+6+12=4\vec c\cdot\vec w = 2(-7)+3(2)+4(3) = -14+6+12 = 4

(a⃗+b⃗)⋅w⃗=±(c⃗⋅w⃗)=±4(\vec a+\vec b)\cdot\vec w = \pm(\vec c\cdot\vec w) = \pm 4

4⇒option (3)\boxed{4 \Rightarrow \text{option (3)}}

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