Let {a}=a_i {i}+a₂ {j}+a₃ {k} and {b}=b₁ {i}+b₂ {j}+b₃ {k} \;\; be two vectors such that |{a}|=1 ; {a}· {b}=2 and |{b}|=4. If {c}=2({a}× {b})-3 {b}, then the angle between {b} and {c} is equal to : [30-Jan-2024 Shift 1]

Correct Answer is : cos⁡−1(−32)^{-1} ≤ft( -{{3}}{2} )cos−1(−23)

Correct Answer is : cos⁡−1(−32)^{-1} ≤ft( -{{3}}{2} )cos−1(−23)

Test Index

Vector Algebra Part 3

Section: Mathematics

Question : 69 of 72

Marks: +1, -0

Let

\vec{a}=a_i \hat{i}+a_2 \hat{j}+a_3 \hat{k}

\vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k} \;\;

|\vec{a}|=1 ; \vec{a}\cdot \vec{b}=2

|\vec{b}|=4

\vec{c}=2(\vec{a}\times \vec{b})-3 \vec{b}

\vec{b}

\vec{c}

Solution: 👈: Video Solution

Given

|\vec{a}|=1,|\vec{b}|=4, \vec{a}\cdot \vec{b}=2

\vec{c}=2(\vec{a}\times \vec{b})-3 \vec{b}

Dot product with

\vec{a}

on both sides

\vec{c}\cdot \vec{a}=-6

........(1)

Dot product with

\vec{b}

on both sides

\;\vec{b}\cdot \vec{c}=-48 \;\; \ldots (2)

\;\vec{c}\cdot \vec{c}=4|\vec{a}\times \vec{b}|^2+9|\vec{b}|^2

\;|\vec{c}|^2=4 \left[ |a|^2|b|^2-(a\cdot \vec{b})^2 \right]

+9|\vec{b}|^2

\;|\vec{c}|^2=4 [(1)(4)^2-(4)] +9(16)

\;|\vec{c}|^2=4[12]+144

\;|\vec{c}|^2=48+144

\;|\vec{c}|^2=192

\; \therefore \cos \theta = \frac{\vec{b}\cdot \vec{c}}{|\vec{b}| |\vec{c}|}

\; \therefore \cos \theta = \frac{-48}{\sqrt{192}\cdot 4}

\; \therefore \cos \theta = \frac{-48}{8 \sqrt{3}\cdot 4}

\; \therefore \cos \theta = \frac{-3}{2 \sqrt{3}}

\; \therefore \cos \theta = \frac{-\sqrt{3}}{2}

\Rightarrow \theta = \cos^{-1} \left( \frac{-\sqrt{3}}{2} \right)

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