The dot products of a vector with the vector $(\hat{i}+\hat{j}-3 \hat{k}),(\hat{i}+3 \hat{j}-2 \hat{k})$ and $(2 \hat{i}+\hat{j}+4 \hat{k})$ are 0,5 and 8 respectively. Find the vector.

Home » The dot products of a vector with the vector and are 0,5 and 8 respectively. Find the vector.

CBSE | Class 12 | Maths | RS Aggarwal | Scalar, or Dot, Product of Vectors

The dot products of a vector with the vector $(\hat{i}+\hat{j}-3 \hat{k}),(\hat{i}+3 \hat{j}-2 \hat{k})$ and $(2 \hat{i}+\hat{j}+4 \hat{k})$ are 0,5 and 8 respectively. Find the vector.

Solution:

$\begin{array}{l} \vec{v}=\mathrm{v} 1 \hat{\imath}+\mathrm{v} 2 \hat{\jmath}+\mathrm{v}_{3} \hat{k} \\ \vec{a}=\hat{\imath}+3 \hat{\jmath}-2 \hat{k} \\ \vec{b}=\hat{\imath}+\hat{\jmath}+\hat{k} \end{array}$
$\vec{c}=2 \hat{\imath}+\hat{\jmath}+4 \hat{k}$
According to the question $\vec{v} \cdot \vec{a}=0, \vec{v} \cdot \vec{b}=5, \vec{v} \cdot \vec{c}=8$
$\begin{array}{l} \mathrm{v} 1+\mathrm{v} 2-3 \mathrm{v} 3=0------I \\ \mathrm{v} 1+3 \mathrm{v} 2-2 \mathrm{v}_{3}=5-------II \\ 2 \mathrm{v} 1+\mathrm{v} 2+4 \mathrm{v}_{3}=8-------III \end{array}$
Using system of linear equations we can simultaneously solve I, II, III to get $\mathrm{v}_{1}=1, \mathrm{v}_{2}=2, \mathrm{v}_{3}=1$ and hence $\vec{v}=\hat{\imath}+2 \hat{\jmath}+\hat{k}$