Let $\bar{a}= \hat{i}+2\hat{j}-\hat{k}$ and $\bar{b}= \hat{i}+\hat{j}-\hat{k}$ be two vectors. If $\bar{c}$ is a vector such that $\bar{b}\times \bar{c}= \bar{b}\times \bar{a}$ and $\bar{c}\cdot \bar{a}=0$ , then $\bar{c}\cdot \bar{b}$ is

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Algebra 2

Transformations of absolute value functions

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Q. Let

\bar{a}= \hat{i}+2\hat{j}-\hat{k}

and

\bar{b}= \hat{i}+\hat{j}-\hat{k}

be two vectors. If

\bar{c}

is a vector such that

\bar{b}\times \bar{c}= \bar{b}\times \bar{a}

and

\bar{c}\cdot \bar{a}=0

, then

\bar{c}\cdot \bar{b}

Identify Cross Product Equation: Identify the cross product equation. $\bar{b}\times \bar{c}= \bar{b}\times \bar{a}$
Calculate Cross Product: Calculate the cross product $\bar{b}\times \bar{a}$ . $\bar{b}\times \bar{a} = (\hat{i}+ \hat{j}- \hat{k})\times(\hat{i}+2 \hat{j}- \hat{k})$ $\bar{b}\times \bar{a} = \hat{i}\times \hat{i} + \hat{i}\times2 \hat{j} + \hat{i}\times(- \hat{k}) + \hat{j}\times \hat{i} + \hat{j}\times2 \hat{j} + \hat{j}\times(- \hat{k}) - \hat{k}\times \hat{i} - \hat{k}\times2 \hat{j} - \hat{k}\times(- \hat{k})$ $\bar{b}\times \bar{a} = 0 + 2 \hat{k} - \hat{j} - \hat{j} + 0 + \hat{i} + \hat{i} - 2 \hat{k} + 0$ $\bar{b}\times \bar{a} = 2 \hat{i} - 2 \hat{j}$
Find Cross Product Equality: Since $\bar{b}\times \bar{c} = \bar{b}\times \bar{a}$ , we have $\bar{b}\times \bar{c} = 2\hat{i} - 2\hat{j}.$
Identify Dot Product Equation: Identify the dot product equation. $\bar{c} \cdot \bar{a} = 0$
Use Dot Product Equation: Use the dot product equation to find components of $\bar{c}$ . Let $\bar{c} = x \hat{i} + y \hat{j} + z \hat{k}$ . $\bar{c}\cdot \bar{a} = (x \hat{i} + y \hat{j} + z \hat{k})\cdot(\hat{i}+2 \hat{j}- \hat{k})$ $\bar{c}\cdot \bar{a} = x\cdot 1 + y\cdot 2 + z\cdot (-1)$ $\bar{c}\cdot \bar{a} = x + 2y - z$ Since $\bar{c}\cdot \bar{a}=0$ , we have $x + 2y - z = 0$ .
Calculate Dot Product of $c$ and $b$ : Now, calculate $\bar{c} \cdot \bar{b}$ using the components of $\bar{c}$ . $\bar{c} \cdot \bar{b} = (x \hat{i} + y \hat{j} + z \hat{k})\cdot(\hat{i}+ \hat{j}- \hat{k})$ $\bar{c} \cdot \bar{b} = x\cdot1 + y\cdot1 + z\cdot(-1)$ $\bar{c} \cdot \bar{b} = x + y - z$
Calculate Dot Product of $c$ and $b$ : Now, calculate $\bar{c}\cdot \bar{b}$ using the components of $\bar{c}$ . $\bar{c}\cdot \bar{b} = (x \hat{i} + y \hat{j} + z \hat{k})\cdot(\hat{i}+ \hat{j}- \hat{k})$ $\bar{c}\cdot \bar{b} = x\cdot 1 + y\cdot 1 + z\cdot (-1)$ $\bar{c}\cdot \bar{b} = x + y - z$ We know $\bar{b}\times \bar{c} = 2 \hat{i} - 2 \hat{j}$ , which means the $i$ and $j$ components of $\bar{c}$ must be such that when crossed with $b$ $1$ , they give $b$ $2$ . However, we made a mistake in the cross product calculation earlier. Let's correct it.