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Algebra of Vectors — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Vectors5 Marks
Question
If the points with position vectors $10\hat{\text{i}}+3\hat{\text{j}},\ 12\hat{\text{i}}-5\hat{\text{j}}$ and $\text{a}\hat{\text{i}}+11\hat{\text{j}}$ are collinear, find the value of a.
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Answer

Let A, B, C be the points with position vectors $10\hat{\text{i}}+3\hat{\text{j}},\ 12\hat{\text{i}}-5\hat{\text{j}}$ and $\text{a}\hat{\text{i}}+11\hat{\text{j}}$. Then, $\overrightarrow{\text{AB}}=$ Position vector of B - Position vector of A $=12\hat{\text{i}}-5\hat{\text{j}}-10\hat{\text{i}}-3\hat{\text{j}}$ $=2\hat{\text{i}}-8\hat{\text{j}}$ $\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B $=\text{a}\hat{\text{i}}+11\hat{\text{j}}-12\hat{\text{i}}+5\hat{\text{j}}$ $=(\text{a}-12)\hat{\text{i}}+16\hat{\text{j}}$ Since, A, B, and C are collinear. $\therefore\ \overrightarrow{\text{AB}}=\lambda\overrightarrow{\text{BC}}$ $\Rightarrow2\hat{\text{i}}-8\hat{\text{j}}=\lambda(\text{a}-12)\hat{\text{i}}+\lambda16\hat{\text{j}}$ $\Rightarrow2=\lambda(\text{a}-12),\ -8=\lambda16$ $\Rightarrow2=\lambda(\text{a}-12),\ \lambda=-\frac{1}2$ $\Rightarrow2=-\frac{1}2(\text{a}-12)$ $\Rightarrow-\text{a}+12=4$ $\Rightarrow\text{a}=8$

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