$$ \begin{aligned} & \text { Given, } A(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}) \text { and } \\ & B(13 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+9 \hat{\mathbf{k}}) \\ & \qquad A B=\sqrt{10^2+5^2+10^2}=15 \end{aligned} $$

TS EAMCET 2020 (Online) 14th September Morning Shift Mathematics - Vector Algebra Question 8 English Explanation

$$ \begin{aligned} &\begin{aligned} A C & =9, D A=6 \\ \because C & =\frac{9(13 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+9 \hat{\mathbf{k}})+6(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})}{9+6} \\ C & =\frac{135 \hat{\mathbf{i}}-30 \hat{\mathbf{j}}+75 \hat{\mathbf{k}}}{15}=9 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}} \end{aligned}\\ &\text { Let position vector of } D \text { is } x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}} \end{aligned} $$

$$ \therefore 3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}=\frac{6 \times(9 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})+9(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}})}{6+9} $$

$$ \Rightarrow \quad 3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}=\frac{2(9 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})+3(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}})}{5} $$

$$ \begin{aligned} & \Rightarrow \quad 15 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-5 \hat{\mathbf{k}} \\ & \quad=(18+3 x) \hat{\mathbf{i}}+(-4+3 y) \hat{\mathbf{j}}+(10+3 z) \hat{\mathbf{k}} \\ & \quad \therefore \quad 15=18+3 x \Rightarrow x=-1 \\ & \quad 5=-4+3 y \Rightarrow y=3 \\ & \text { and }-5=10+3 z \Rightarrow \quad z=-5 \\ & \therefore \text { Position vector of } D \text { is }-\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}} \end{aligned} $$

Vector Algebra

Continue this paper

Similar Vector Algebra PYQs

More practice from this question