Question

If ${\stackrel{\mathbf{\rightharpoonup }}{\mathbf{d}}}_{\mathbf{1}}\mathbf{=}\mathbf{3}\hat{\mathbf{i}}\mathbf{-}\mathbf{2}\hat{\mathbf{j}}\mathbf{+}\mathbf{4}\hat{\mathbf{k}}$and ${\stackrel{\mathbf{\rightharpoonup }}{\mathbf{d}}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{5}\hat{\mathbf{i}}\mathbf{+}\mathbf{2}\hat{\mathbf{j}}\mathbf{-}\hat{\mathbf{k}}$then what is $\left({\stackrel{\rightharpoonup }{d}}_1+{\stackrel{\rightharpoonup }{d}}_2\right)\mathbf{.}\left({\stackrel{\rightharpoonup }{d}}_1\times 4{\stackrel{\rightharpoonup }{d}}_2\right)$?

Short answer

The product,$\mathbf{(}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{d}}}_{\mathbf{1}}\mathbf{+}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{d}}}_{\mathbf{2}}\mathbf{)}\mathbf{.}\mathbf{(}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{d}}}_{\mathbf{1}}\mathbf{\times }\mathbf{4}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{d}}}_{\mathbf{2}}\mathbf{)}$is equal to 0.

Step-by-step solution

Vector operations

The dot product of perpendicular vectors is equal to zero. The cross-product of two vectors results in a vector that is perpendicular to both vectors.

Find the value for $\left({\stackrel{\rightharpoonup }{d}}_1+{\stackrel{\rightharpoonup }{d}}_2\right)$and $\left({\stackrel{\rightharpoonup }{d}}_1\times 4{\stackrel{\rightharpoonup }{d}}_2\right)$separately and then take the dot product of them to find the required answer.

The given vectors are,

$$\begin{gathered} \stackrel{\rightharpoonup }{{\mathrm{d}}_1}=3\hat{\mathrm{i}}-2\hat{\mathrm{j}}+4\hat{\mathrm{k}} \\ {\stackrel{\rightharpoonup }{\mathrm{d}}}_2=-5\hat{\mathrm{j}}+2\hat{\mathrm{j}}-\hat{\mathrm{k}} \end{gathered}$$

Calculating the cross and dot product

${\stackrel{\rightharpoonup }{\mathrm{d}}}_1+{\stackrel{\rightharpoonup }{\mathrm{d}}}_2$will lie in the same plane i.e., in the plane of ${\stackrel{\rightharpoonup }{\mathrm{d}}}_1\mathrm{and}{\stackrel{\rightharpoonup }{\mathrm{d}}}_2$.

${\stackrel{\rightharpoonup }{\mathrm{d}}}_1\times 4{\stackrel{\rightharpoonup }{\mathrm{d}}}_2$ will lie in the plane which is perpendicular to the plane of ${\stackrel{\rightharpoonup }{\mathrm{d}}}_1\mathrm{and}{\stackrel{\rightharpoonup }{\mathrm{d}}}_2$.

As they are perpendicular to each other the dot product of ${\stackrel{\rightharpoonup }{\mathrm{d}}}_1+{\stackrel{\rightharpoonup }{\mathrm{d}}}_2$and${\stackrel{\rightharpoonup }{\mathrm{d}}}_1\times 4{\stackrel{\rightharpoonup }{\mathrm{d}}}_2$will be zero.

role="math" localid="1656309525133" $\left({\stackrel{\rightharpoonup }{\mathrm{d}}}_1+{\stackrel{\rightharpoonup }{\mathrm{d}}}_2\right)$.$\left({\stackrel{\rightharpoonup }{\mathrm{d}}}_1\times 4{\stackrel{\rightharpoonup }{\mathrm{d}}}_2\right)$=0

Thus, the dot product of $\left({\stackrel{\rightharpoonup }{\mathrm{d}}}_1+{\stackrel{\rightharpoonup }{\mathrm{d}}}_2\right)$and $\left({\stackrel{\rightharpoonup }{\mathrm{d}}}_1\times 4{\stackrel{\rightharpoonup }{\mathrm{d}}}_2\right)$is equal to 0.