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A vector $\vec{b}$ is collinear with the vector $\vec{a}=(2,1,-1)$ and
satisfies the condition $\vec{a} \cdot \vec{b}=3 .$ What is $\vec{b}$ equal to?
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satisfies the condition $\vec{a} \cdot \vec{b}=3 .$ What is $\vec{b}$ equal to?
Solution:
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Verified Answer
The correct answer is:
$(1,1 / 2,-1 / 2)$
Let $\vec{b}=x \vec{i}+y \vec{j}+z \vec{k}$
Since, $\vec{b}$ is collinear with vetor $\vec{a}$
therefore $\vec{a}=k \vec{b}$ where $k$ is a scalar.
Given $\vec{a}=(2,1,-1)$
$\therefore \quad(2,1,-1)=k(x, y, z)$
$\Rightarrow x=\frac{2}{k}, y=\frac{1}{k}, z=\frac{-1}{k}$
Also, $\vec{a} \cdot \vec{b}=3$
$\Rightarrow 2 x+y-z=3$
$\Rightarrow \quad 2\left(\frac{2}{k}\right)+\frac{1}{k}+\frac{1}{k}=3$
$\Rightarrow \frac{4}{k}+\frac{1}{k}+\frac{1}{k}=3$
$\Rightarrow \frac{6}{k}=3 \Rightarrow k=2$
$\therefore \quad x=1, y=\frac{1}{2}$ and $z=\frac{-1}{2}$
Hence $\vec{b}=\left(1, \frac{1}{2}, \frac{-1}{2}\right)$
Since, $\vec{b}$ is collinear with vetor $\vec{a}$
therefore $\vec{a}=k \vec{b}$ where $k$ is a scalar.
Given $\vec{a}=(2,1,-1)$
$\therefore \quad(2,1,-1)=k(x, y, z)$
$\Rightarrow x=\frac{2}{k}, y=\frac{1}{k}, z=\frac{-1}{k}$
Also, $\vec{a} \cdot \vec{b}=3$
$\Rightarrow 2 x+y-z=3$
$\Rightarrow \quad 2\left(\frac{2}{k}\right)+\frac{1}{k}+\frac{1}{k}=3$
$\Rightarrow \frac{4}{k}+\frac{1}{k}+\frac{1}{k}=3$
$\Rightarrow \frac{6}{k}=3 \Rightarrow k=2$
$\therefore \quad x=1, y=\frac{1}{2}$ and $z=\frac{-1}{2}$
Hence $\vec{b}=\left(1, \frac{1}{2}, \frac{-1}{2}\right)$
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