Question

Here are three vectors in meters:

$\overrightarrow{{\mathbf{d}}_{\mathbf{1}}}\mathbf{=}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\hat{\mathbf{j}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}$
localid="1654741448647" $\overrightarrow{{\mathbf{d}}_2}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}$

localid="1654741501014" $\overrightarrow{{\mathbf{d}}_3}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}$

What result from (a)localid="1654740492420" ${}^{\overrightarrow{{\mathbf{d}}_{\mathbf{1}}}\mathbf{.}\mathbf{(}\overrightarrow{{\mathbf{d}}_{\mathbf{2}}}\mathbf{}\mathbf{+}\mathbf{}\overrightarrow{{\mathbf{d}}_{\mathbf{3}}}\mathbf{)}}$, (b) ${}^{\overrightarrow{{\mathbf{d}}_{\mathbf{1}}}\mathbf{.}\mathbf{(}\overrightarrow{{\mathbf{d}}_{\mathbf{2}}}\mathbf{}\mathbf{\times }\mathbf{}\overrightarrow{{\mathbf{d}}_{\mathbf{3}}}\mathbf{)}}$, and (c) localid="1654740727403" ${}^{{\mathbf{d}}_{\mathbf{1}}\mathbf{\times }\mathbf{}\mathbf{(}\mathbf{}{\mathbf{d}}_{\mathbf{2}}\mathbf{+}{\mathbf{d}}_{\mathbf{3}\mathbf{}}\mathbf{)}}$ ?

Short answer

(a) ${}^{\overrightarrow{{\mathbf{d}}_{\mathbf{1}}}\mathbf{.}\mathbf{(}\overrightarrow{{\mathbf{d}}_{\mathbf{2}}}\mathbf{}\mathbf{+}\mathbf{}\overrightarrow{{\mathbf{d}}_{\mathbf{3}}}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}{\mathbf{m}}^{\mathbf{2}}}$

(b) ${}^{\overrightarrow{{\mathbf{d}}_{\mathbf{1}}}\mathbf{.}\mathbf{}\mathbf{(}\overrightarrow{{\mathbf{d}}_{\mathbf{2}}}\mathbf{}\mathbf{\times }\mathbf{}\overrightarrow{{\mathbf{d}}_{\mathbf{3}}}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{52}\mathbf{.}\mathbf{0}\mathbf{}{\mathbf{m}}^{\mathbf{3}}}$

(c) ${}^{\overrightarrow{{\mathbf{d}}_{\mathbf{1}}}\mathbf{\times }\mathbf{}\mathbf{(}\overrightarrow{{\mathbf{d}}_{\mathbf{2}}}\mathbf{}\mathbf{+}\mathbf{}\overrightarrow{{\mathbf{d}}_{\mathbf{3}}}\mathbf{)}\mathbf{}\mathbf{=}\left(11.0m^2\right)\hat{\mathbf{i}}\mathbf{}\mathbf{+}\left(9.0m^2\right)\hat{\mathbf{j}}\mathbf{}\mathbf{+}\left(3.0m^2\right)\hat{\mathbf{k}}}$

Step-by-step solution

Given data

Three vectors are given as:

$\overrightarrow{{\mathbf{d}}_{\mathbf{1}}}\mathbf{=}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}$

role="math" localid="1654741486353" $\overrightarrow{{\mathbf{d}}_2}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}$

$\overrightarrow{{\mathbf{d}}_3}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}$

Understanding the vector operations

This problem refers to vector operations that include vector addition, dot product and cross product. Dot product is a scalar quantity whereas cross product is a vector quantity.

The expression for the vector product is given as follows:

$\left(\overrightarrow{a}\times \overrightarrow{b}\right)\mathbf{}\mathbf{=}\left(a_y{b}_z-b_y{a}_z\right)\hat{\mathbf{i}}\mathbf{+}\left(a_z{b}_x-b_z{a}_x\right)\hat{\mathbf{j}}\mathbf{+}\left(a_x{b}_y-b_y{a}_x\right)\hat{\mathbf{k}}$ … (i)

The expression for the dot product is given as follows:

$\left(\overrightarrow{a}.\overrightarrow{b}\right)\mathbf{}\mathbf{=}\left(a_x{b}_x+a_y{b}_y+a_z{b}_z\right)$ … (ii)

(a) Determination of d→1.(d→2+d→3)

First, addition of ${}^{\overrightarrow{{\mathbf{d}}_{\mathbf{2}}}}$and $\overrightarrow{{}^{{\mathbf{d}}_{\mathbf{3}}}}$gives,

role="math" localid="1654743006461" $$\begin{gathered} \mathbf{(}\overrightarrow{{\mathbf{d}}_{\mathbf{2}}}\mathbf{}\mathbf{+}\overrightarrow{{\mathbf{d}}_{\mathbf{3}}}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{)}\hat{\mathbf{i}}\mathbf{+}\mathbf{(}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{)}\hat{\mathbf{j}}\mathbf{+}\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{)}\hat{\mathbf{k}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{k}} \end{gathered}$$

Now, the dot product of $\overrightarrow{{}^{{\mathbf{d}}_{\mathbf{1}}}}$gives,

$$\begin{gathered} \overrightarrow{{\mathbf{d}}_{\mathbf{1}}}\mathbf{.}\mathbf{(}\overrightarrow{{\mathbf{d}}_{\mathbf{2}}}\mathbf{}\mathbf{+}\overrightarrow{{\mathbf{d}}_{\mathbf{3}}}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}\mathbf{)}\mathbf{}\mathbf{.}\mathbf{}\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{+}\mathbf{6}\mathbf{.}\mathbf{0} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}{\mathbf{m}}^{\mathbf{2}} \end{gathered}$$

(b) Determination of d→1.(d→2×d→3)

The cross product of ${}^{\overrightarrow{{\mathbf{d}}_{\mathbf{2}}}}$and ${}^{\overrightarrow{{\mathbf{d}}_{\mathbf{3}}}}$gives,

$\mathbf{(}\overrightarrow{{\mathbf{d}}_{\mathbf{2}}}\mathbf{}\mathbf{\times }\mathbf{}\overrightarrow{{\mathbf{d}}_{\mathbf{3}}}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{-}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{)}\hat{\mathbf{i}}\mathbf{+}\mathbf{(}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{-}\left(-2.0\right)\mathbf{)}\hat{\mathbf{j}}\mathbf{+}\mathbf{(}\left(-6.0\right)\mathbf{-}\left(-8.0\right)\mathbf{)}\hat{\mathbf{k}}$

$\mathbf{(}\overrightarrow{{\mathbf{d}}_{\mathbf{2}}}\mathbf{}\mathbf{\times }\mathbf{}\overrightarrow{{\mathbf{d}}_{\mathbf{3}}}\mathbf{)}\mathbf{=}\mathbf{-}\mathbf{10}\hat{\mathbf{i}}\mathbf{}\mathbf{+}\mathbf{6}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}$

Now, the dot product of $\overrightarrow{{}^{{\mathbf{d}}_{\mathbf{1}}}}$gives,

role="math" localid="1654743321424" $$\begin{gathered} \overrightarrow{{\mathbf{d}}_{\mathbf{1}}}\mathbf{.}\mathbf{(}\overrightarrow{{\mathbf{d}}_{\mathbf{2}}}\mathbf{}\mathbf{\times }\overrightarrow{{\mathbf{d}}_{\mathbf{3}}}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}\mathbf{)}\mathbf{}\mathbf{.}\mathbf{}\mathbf{(}\mathbf{-}\mathbf{10}\hat{\mathbf{i}}\mathbf{}\mathbf{+}\mathbf{6}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{+}\mathbf{18}\mathbf{.}\mathbf{0}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{0} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{52}\mathbf{.}\mathbf{0}\mathbf{}{\mathbf{m}}^3 \end{gathered}$$

(c) Determination of d→1×(d→2+d→3)

The addition of ${}^{\overrightarrow{{\mathbf{d}}_{\mathbf{2}}}}$and $\overrightarrow{{}^{{\mathbf{d}}_{\mathbf{3}}}}$gives ,

$\mathbf{(}\overrightarrow{{\mathbf{d}}_{\mathbf{2}}}\mathbf{}\mathbf{+}\overrightarrow{{\mathbf{d}}_{\mathbf{3}}}\mathbf{)}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}$

Now, now the cross product of ${}^{\overrightarrow{{\mathbf{d}}_{\mathbf{1}}}}$gives,

localid="1660891291873" $$\begin{gathered} \overrightarrow{{\mathbf{d}}_{\mathbf{1}}}\times \mathbf{(}\overrightarrow{{\mathbf{d}}_{\mathbf{2}}}\mathbf{}\mathbf{+}\overrightarrow{{\mathbf{d}}_{\mathbf{3}}}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}\mathbf{)}\mathbf{}\mathbf{\times }\mathbf{}\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{(}\mathbf{11}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{)}\hat{\mathbf{i}}\mathbf{+}\mathbf{(}\mathbf{9}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{)}\hat{\mathbf{j}}\mathbf{+}\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{)}\hat{\mathbf{k}} \end{gathered}$$