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Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry5 Marks
Question
$\text{If }\overrightarrow{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{7k}};\text{ and }\overrightarrow{b}=\hat{\text{5i}}-\hat{\text{j}}+\lambda\hat{\text{k}},$ then find the value of $\lambda,$ so that $\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-\overrightarrow{b}$ are perpendicular vectors.
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Answer

Here $\text{If }\overrightarrow{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{7k}};\text{ }\overrightarrow{b}=\hat{\text{5i}}-\hat{\text{j}}+\lambda\hat{\text{k}},$$\therefore \overrightarrow{a}+\overrightarrow{b}=6\hat{\text{i}}-2\hat{\text{j}}+\text{(7}+\lambda)\hat{\text{k}};\text{ } \overrightarrow{a}-\overrightarrow{b}=-4\hat{\text{i}}+\text{(7}-\lambda)\hat{\text{k}}$
$\therefore\text{ }\overrightarrow{(a}+\overrightarrow{b)}\text{ is perpendicular to }\overrightarrow{(a}-\overrightarrow{b)}$
$\Rightarrow \text{ }\overrightarrow{(a}+\overrightarrow{b)}.\overrightarrow{(a}-\overrightarrow{b)}=0\text{ }\Rightarrow \text{ -24 + (7}+\lambda).\text{(7} - \lambda)=0$
$\Rightarrow$ –24 + 49 – $\lambda^2 $ = 0 $\Rightarrow$ $\lambda^2 $ = 25
$\Rightarrow$ $\lambda $ = ± 5.

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