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Vector Algebra Part 3

Section: Mathematics

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Question : 47 of 72

Marks: +1, -0

Let

\vec{a}=4\hat{i}-\hat{j}+\hat{k}, \vec{b}=11\hat{i}-\hat{j}+\hat{k}

\vec{c}

be a vector such that

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

(2 \vec{a}+3 \vec{b}) \cdot \vec{c}=1670

|\vec{c}|^2

[8 Apr 2024 Shift 2]

1627
1618
1600
1609

Solution: 👈: Video Solution

\;(\vec{a}+\vec{b}) \times \vec{c}-\vec{c} \times(-2 \vec{a}+3 \vec{b})=0

\;(\vec{a}+\vec{b}) \times \vec{c}+(-2 \vec{a}+3 \vec{b}) \times \vec{c}=0

\;\Rightarrow(\vec{a}+\vec{b})-2 \vec{a}+3 \vec{b}) \times \vec{c}=0

\;\Rightarrow \vec{c}=\lambda(4 \vec{b}-\vec{a})

\;\Rightarrow =\lambda(44 \hat{i}-4 \hat{j}+4 \hat{k}-4 \hat{i}+\hat{j}-\hat{k})

\;=\lambda(40 \hat{i}-3 \hat{j}+3 \hat{k})

Now

\;(8 \hat{i}-2 \hat{j}+2 \hat{k}+33 \hat{i}-3 \hat{j}+3 \hat{k}) \cdot \lambda(40 \hat{i}-3 \hat{j}+3 \hat{k})=1670

\;\Rightarrow(41 \hat{i}-5 \hat{j}+5 \hat{k}) \cdot(40 \hat{i}-3 \hat{j}+3 \hat{k}) \times \lambda=1670)

\;\Rightarrow(1640+15+15) \lambda=1670 \Rightarrow \lambda=1

\;\;\text{ so }\;\vec{c}=40 \hat{i}-3 \hat{j}-3 \hat{k}

\;\Rightarrow|\vec{c}|^2=1600+9+9=1618

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