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Found in: Page 56

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# Which of the following are correct (meaningful) vector expressions? What is wrong with any incorrect expression?${\mathbf{\left(}}{\mathbf{a}}{\mathbf{\right)}}{\mathbf{ }}\stackrel{\mathbf{\to }}{\mathbf{A}}{\mathbf{}}{\mathbf{\cdot }}{\mathbf{\left(}}\stackrel{\mathbf{\to }}{\mathbf{B}}{\mathbf{\cdot }}\stackrel{\mathbf{\to }}{\mathbf{C}}{\mathbf{}}{\mathbf{\right)}}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{\left(}}{\mathbf{f}}{\mathbf{\right)}}{\mathbf{ }}\stackrel{\mathbf{\to }}{\mathbf{A}}{\mathbf{}}{\mathbf{+}}{\mathbf{\left(}}{\mathbf{B}}{\mathbf{}}{\mathbf{⃗}}{\mathbf{×}}{\mathbf{C}}{\mathbf{}}{\mathbf{⃗}}{\mathbf{}}{\mathbf{\right)}}\phantom{\rule{0ex}{0ex}}{\mathbf{\left(}}{\mathbf{b}}{\mathbf{\right)}}{\mathbf{}}\stackrel{\mathbf{\to }}{\mathbf{A}}{\mathbf{}}{\mathbf{×}}{\mathbf{\left(}}\stackrel{\mathbf{\to }}{\mathbf{B}}{\mathbf{\cdot }}\stackrel{\mathbf{\to }}{\mathbf{C}}{\mathbf{}}{\mathbf{\right)}}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{\left(}}{\mathbf{g}}{\mathbf{\right)}}{\mathbf{ }}{\mathbf{5}}{\mathbf{+}}\stackrel{\mathbf{\to }}{\mathbf{A}}{\mathbf{}}\phantom{\rule{0ex}{0ex}}{\mathbf{\left(}}{\mathbf{c}}{\mathbf{\right)}}{\mathbf{}}\stackrel{\mathbf{\to }}{\mathbf{}\mathbf{A}}{\mathbf{\cdot }}{\mathbf{\left(}}\stackrel{\mathbf{\to }}{\mathbf{B}}{\mathbf{×}}\stackrel{\mathbf{\to }}{\mathbf{C}}{\mathbf{}}{\mathbf{\right)}}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{\left(}}{\mathbf{h}}{\mathbf{\right)}}{\mathbf{ }}{\mathbf{5}}{\mathbf{+}}{\mathbf{\left(}}\stackrel{\mathbf{\to }}{\mathbf{B}}{\mathbf{\cdot }}\stackrel{\mathbf{\to }}{\mathbf{C}}{\mathbf{\right)}}{\mathbf{ }}{\mathbf{}}\phantom{\rule{0ex}{0ex}}{\mathbf{\left(}}{\mathbf{d}}{\mathbf{\right)}}{\mathbf{}}{\mathbf{}}{\mathbf{}}\stackrel{\mathbf{\to }}{\mathbf{A}}{\mathbf{}}{\mathbf{×}}{\mathbf{\left(}}\stackrel{\mathbf{\to }}{\mathbf{B}}{\mathbf{×}}\stackrel{\mathbf{\to }}{\mathbf{C}}{\mathbf{}}{\mathbf{\right)}}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{ }}{\mathbf{ }}{\mathbf{\left(}}{\mathbf{i}}{\mathbf{\right)}}{\mathbf{ }}{\mathbf{5}}{\mathbf{+}}{\mathbf{\left(}}\stackrel{\mathbf{\to }}{\mathbf{B}}{\mathbf{×}}\stackrel{\mathbf{\to }}{\mathbf{C}}{\mathbf{}}{\mathbf{}}{\mathbf{\right)}}\phantom{\rule{0ex}{0ex}}{\mathbf{}}{\mathbf{\left(}}{\mathbf{e}}{\mathbf{\right)}}{\mathbf{ }}\stackrel{\mathbf{\to }}{\mathbf{A}}{\mathbf{}}{\mathbf{+}}{\mathbf{\left(}}\stackrel{\mathbf{\to }}{\mathbf{B}}{\mathbf{\cdot }}\stackrel{\mathbf{\to }}{\mathbf{C}}{\mathbf{}}{\mathbf{\right)}}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{ }}{\mathbf{\left(}}{\mathbf{j}}{\mathbf{\right)}}{\mathbf{ }}{\mathbf{\left(}}\stackrel{\mathbf{\to }}{\mathbf{A}}{\mathbf{\cdot }}\stackrel{\mathbf{\to }}{\mathbf{B}}{\mathbf{}}{\mathbf{\right)}}{\mathbf{+}}{\mathbf{\left(}}\stackrel{\mathbf{\to }}{\mathbf{B}}{\mathbf{×}}\stackrel{\mathbf{\to }}{\mathbf{C}}{\mathbf{}}{\mathbf{\right)}}$

Correct Expressions are (c), (d), (f) and (h).

See the step by step solution

## Step 1: Given information

Expressions are given in the problem to verify.

$\left(\mathrm{a}\right) \stackrel{\to }{\mathrm{A}}\cdot \left(\stackrel{\to }{\mathrm{B}}\cdot \stackrel{\to }{\mathrm{C}}\right) \left(\mathrm{f}\right) \stackrel{\to }{\mathrm{A}}+\left(\mathrm{B}⃗×\mathrm{C}⃗\right)\phantom{\rule{0ex}{0ex}}\left(\mathrm{b}\right)\stackrel{\to }{\mathrm{A}}×\left(\stackrel{\to }{\mathrm{B}}\cdot \stackrel{\to }{\mathrm{C}}\right) \left(\mathrm{g}\right) 5+\stackrel{\to }{\mathrm{A}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{c}\right)\stackrel{\to }{\mathrm{A}}\cdot \left(\stackrel{\to }{\mathrm{B}}×\stackrel{\to }{\mathrm{C}}\right) \left(\mathrm{h}\right) 5+\left(\stackrel{\to }{\mathrm{B}}\cdot \stackrel{\to }{\mathrm{C}}\right) \phantom{\rule{0ex}{0ex}}\left(\mathrm{d}\right)\stackrel{\to }{\mathrm{A}}×\left(\stackrel{\to }{\mathrm{B}}×\stackrel{\to }{\mathrm{C}}\right) \left(\mathrm{i}\right) 5+\left(\stackrel{\to }{\mathrm{B}}×\stackrel{\to }{\mathrm{C}}\right)\phantom{\rule{0ex}{0ex}}\left(\mathrm{e}\right) \stackrel{\to }{\mathrm{A}}+\left(\stackrel{\to }{\mathrm{B}}\cdot \stackrel{\to }{\mathrm{C}}\right) \left(\mathrm{j}\right) \left(\stackrel{\to }{\mathrm{A}}\cdot \stackrel{\to }{\mathrm{B}}\right)+\left(\stackrel{\to }{\mathrm{B}}×\stackrel{\to }{\mathrm{C}}\right)$

## Step 2: To understand the concept

The scalar product of two vectors is a scalar quantity. If the angle between the vector $\stackrel{\mathbf{\to }}{\mathbf{A}}{\mathbf{}}$ and vector role="math" localid="1657009109003" $\stackrel{\mathbf{\to }}{\mathbf{B}}$ is ${\mathbf{\theta }}$ , the scalar product can be written as,

$\stackrel{\mathbf{\to }}{\mathbf{A}}{\mathbf{\cdot }}\stackrel{\mathbf{\to }}{\mathbf{B}}{\mathbf{=}}{\mathbf{AB}}{\mathbf{}}{\mathbf{cos\theta }}$

It is also called the dot product. The scalar product can be found for vectors only. We cannot calculate the scalar product between the two scalar quantities or one scalar and the other vector quantity.

The vector product of two vectors is a vector quantity. If the angle between the vector $\stackrel{\mathbf{\to }}{\mathbf{A}}{\mathbf{}}$ and vector $\stackrel{\mathbf{\to }}{\mathbf{B}}$ is ${\mathbf{\theta }}$, the vector product can be written as,

$\stackrel{\mathbf{\to }}{\mathbf{A}}{\mathbf{×}}\stackrel{\mathbf{\to }}{\mathbf{B}}{\mathbf{=}}{\mathbf{\left(}}{{\mathbf{A}}}_{{\mathbf{y}}}{{\mathbf{B}}}_{{\mathbf{z}}}{\mathbf{-}}{{\mathbf{B}}}_{{\mathbf{y}}}{{\mathbf{A}}}_{{\mathbf{z}}}{\mathbf{\right)}}\stackrel{\mathbf{^}}{\mathbf{i}}{\mathbf{+}}{\mathbf{\left(}}{{\mathbf{A}}}_{{\mathbf{z}}}{{\mathbf{B}}}_{{\mathbf{x}}}{\mathbf{-}}{{\mathbf{B}}}_{{\mathbf{z}}}{{\mathbf{A}}}_{{\mathbf{x}}}{\mathbf{\right)}}\stackrel{\mathbf{^}}{\mathbf{j}}{\mathbf{+}}{\mathbf{\left(}}{{\mathbf{A}}}_{{\mathbf{x}}}{{\mathbf{B}}}_{{\mathbf{y}}}{\mathbf{-}}{{\mathbf{B}}}_{{\mathbf{x}}}{{\mathbf{A}}}_{{\mathbf{y}}}{\mathbf{\right)}}\stackrel{\mathbf{^}}{\mathbf{k}}$

The vector product is also called the cross product. We cannot calculate the vector product between the two scalar quantities or one scalar and another vector quantity.

## Step 3: To find the correct vector expressions

We can use the properties of addition and multiplication of a vector.

Also, the concept of vector dot product and vector cross product is useful here.

(a) In this expression, $\left(\stackrel{\to }{\mathrm{B}}·\stackrel{\to }{\mathrm{C}}\right)$ is a dot product, so the results will be the scalar. As explained in the concept, we cannot find the dot product with a scalar, so the vector expression $\stackrel{\to }{\mathrm{A}·}\left(\stackrel{\to }{\mathrm{B}}·\stackrel{\to }{\mathrm{C}}\right)$is wrong.

(b) $\left(\stackrel{\to }{\mathrm{B}}·\stackrel{\to }{\mathrm{C}}\right)$is a scalar. As explained in the concept, we cannot find the cross product of $\stackrel{\to }{\mathrm{A}}$ with scalar $\left(\stackrel{\to }{\mathrm{B}}·\stackrel{\to }{\mathrm{C}}\right)$.

(c) $\left(\stackrel{\to }{\mathrm{B}}×\stackrel{\to }{\mathrm{C}}\right)$ is a vector quantity. As explained in the concept, we can find the dot product of $\stackrel{\to }{\mathrm{A}}$ with $\left(\stackrel{\to }{\mathrm{B}}×\stackrel{\to }{\mathrm{C}}\right)$, as both are vector quantities. Therefore,$\stackrel{\to }{\mathrm{A}}·\left(\stackrel{\to }{\mathrm{B}}×\stackrel{\to }{\mathrm{C}}\right)$ is the right vector expression.

(d) $\left(\stackrel{\to }{\mathrm{B}}×\stackrel{\to }{\mathrm{C}}\right)$is a vector quantity. As explained in the concept, we can find the cross product of $\stackrel{\to }{\mathrm{A}}$ with $\left(\stackrel{\to }{\mathrm{B}}×\stackrel{\to }{\mathrm{C}}\right)$ , as both are vector quantities. Therefore, role="math" localid="1657009831768" $\stackrel{\to }{\mathrm{A}}×\left(\stackrel{\to }{\mathrm{B}}×\stackrel{\to }{\mathrm{C}}\right)$ is the right vector expression.

(e) $\left(\stackrel{\to }{\mathrm{B}}·\stackrel{\to }{\mathrm{C}}\right)$is a scalar quantity because $\left(\stackrel{\to }{\mathrm{B}}·\stackrel{\to }{\mathrm{C}}\right)$is a dot product. We cannot add a scalar to the vector quantity. Therefore, $\stackrel{\to }{\mathrm{A}}$ cannot be added with a scalar $\left(\stackrel{\to }{\mathrm{B}}·\stackrel{\to }{\mathrm{C}}\right)$.

(f) $\left(\stackrel{\to }{\mathrm{B}}×\stackrel{\to }{\mathrm{C}}\right)$is a vector quantity. Therefore, it is possible to add vector $\stackrel{\to }{\mathrm{A}}$ with another vector$\left(\stackrel{\to }{\mathrm{B}}×\stackrel{\to }{\mathrm{C}}\right)$ . Therefore, it is the right vector expression.

(g) We cannot add scaler 5 with vector $\stackrel{\to }{\mathrm{A}}$. Therefore, it is not right expression.

(h) role="math" localid="1657010159376" $\left(\stackrel{\to }{\mathrm{B}}·\stackrel{\to }{\mathrm{C}}\right)$is a scalar quantity. It is possible to add two scalar quantities. Therefore, it is the right vector expression.

(I) We cannot add scalar number 5 with $\left(\stackrel{\to }{\mathrm{B}}×\stackrel{\to }{\mathrm{C}}\right)$ . A cross product of two vectors is a vector. So, we cannot do the addition of a scalar with a vector. Therefore, it is not right expression.

(j) $\left(\stackrel{\to }{\mathrm{A}}\cdot \stackrel{\to }{\mathrm{B}}\right)$ is a scalar, and $\left(\stackrel{\to }{\mathrm{B}}×\stackrel{\to }{\mathrm{C}}\right)$ is a vector. So, we cannot do the addition of scalar with a vector.

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