Question

If $\overrightarrow{{\mathbf{d}}_{\mathbf{1}}}\mathbf{+}\overrightarrow{{\mathbf{d}}_{\mathbf{2}}}\mathbf{=}\mathbf{5}\overrightarrow{{\mathbf{d}}_{\mathbf{3}}}\mathbf{,}\mathbf{}\overrightarrow{{\mathbf{d}}_{\mathbf{1}}}\mathbf{-}\overrightarrow{{\mathbf{d}}_{\mathbf{2}}}\mathbf{=}\mathbf{3}\overrightarrow{{\mathbf{d}}_{\mathbf{3}}}$and$\overrightarrow{{\mathbf{d}}_{\mathbf{3}}}\mathbf{=}\mathbf{2}\hat{\mathbf{i}}\mathbf{+}\mathbf{4}\hat{\mathbf{j}}$unit vector notation (a) $\overrightarrow{{\mathbf{d}}_{\mathbf{1}}}$and (b)$\overrightarrow{{\mathbf{d}}_{\mathbf{2}}}$?

Short answer

(a)$\overrightarrow{d_1}=8\hat{i}+16\hat{j}$

(b)$\overrightarrow{d_3}=2\hat{i}+4\hat{j}$

Step-by-step solution

Given data

The following information is given.

$\overrightarrow{d_1}+\overrightarrow{d_2}=5\overrightarrow{d_3}$ (i)

$\overrightarrow{d_1}-\overrightarrow{d_2}=3\overrightarrow{d_3}$ (ii)

$\overrightarrow{d_3}=2\hat{i}+4\hat{j}$ (iii)

Understanding the concept

Use the vector addition and vector subtraction laws to calculate the values of vector$\overrightarrow{{\mathbf{d}}_{\mathbf{1}}}$and $\overrightarrow{{\mathbf{d}}_2}$.

Determination ofd1 andd2.

To calculate the values of vectors, add equations (i) and (ii) as follows:

$\begin{array}{rcl}\left(\overrightarrow{d_1}+\overrightarrow{d_2}\right)+\left(\overrightarrow{d_1}-\overrightarrow{d_2}\right)& =& 5\overrightarrow{d_3}+3\overrightarrow{d_3}\\ & =& 8\overrightarrow{d_3}\\ 2\overrightarrow{d_1}& =& 8\overrightarrow{d_3}\\ \overrightarrow{d_1}& =& 4\overrightarrow{d_3}\\ & & \end{array}$

From equation (iii),

$\overrightarrow{d_1}+\overrightarrow{d_2}=5\overrightarrow{d_3}$

Substituting the value of $\overrightarrow{d_1}$, we get

$\overrightarrow{d_2}=\overrightarrow{d_3}$

Therefore, the vector in $\overrightarrow{d_2}$unit vector notation is,

$\overrightarrow{d_2}=2\hat{i}+4\hat{j}$

Now, use the value of $\overrightarrow{d_2}$to calculate the $\overrightarrow{d_1}$

$\begin{array}{rcl}\overrightarrow{d_1}& =& 4\times \overrightarrow{d_3}\\ & =& 4\times \left(2\hat{i}+4\hat{j}\right)\\ & =& 8\hat{i}+16\hat{j}\\ & & \end{array}$

Therefore, $\overrightarrow{d_1}=8\hat{i}+16\hat{j}$androle="math" localid="1660886521162" $\overrightarrow{d_2}=2\hat{i}+4\hat{j}$