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MathematicsInteger+3Question 49 of 84

IIT-JEE 2010 · Paper 1, Mathematics question 49

If a⃗\vec a and b⃗\vec b are vectors in space given by a⃗=i^−2j^5\vec a=\dfrac{\hat i-2\hat j}{\sqrt5} and b⃗=2i^+j^+3k^14\vec b=\dfrac{2\hat i+\hat j+3\hat k}{\sqrt{14}}, then the value of (2a⃗+b⃗)⋅[(a⃗×b⃗)×(a⃗−2b⃗)](2\vec a+\vec b)\cdot\left[(\vec a\times\vec b)\times(\vec a-2\vec b)\right] is

Answer and worked solution

Official answer: 5

Facts. ∣a⃗∣=∣b⃗∣=1|\vec a|=|\vec b|=1 and a⃗⋅b⃗=2−2+070=0\vec a\cdot\vec b=\frac{2-2+0}{\sqrt{70}}=0.

Expand the triple product.
(a⃗×b⃗)×(a⃗−2b⃗)=[(a⃗⋅a⃗)b⃗−(a⃗⋅b⃗)a⃗]−2[(a⃗⋅b⃗)b⃗−(b⃗⋅b⃗)a⃗]=b⃗+2a⃗.(\vec a\times\vec b)\times(\vec a-2\vec b)=\left[(\vec a\cdot\vec a)\vec b-(\vec a\cdot\vec b)\vec a\right]-2\left[(\vec a\cdot\vec b)\vec b-(\vec b\cdot\vec b)\vec a\right]=\vec b+2\vec a.

Dot product. (2a⃗+b⃗)⋅(2a⃗+b⃗)=4+1=5(2\vec a+\vec b)\cdot(2\vec a+\vec b)=4+1=5.

Answer: 5

Topics:Scalar and vector triple products
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