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Algebra of Vectors — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Vectors1 Mark
MCQ
If points $\text{A}(60\hat{\text{i}}+3\hat{\text{j}}),\ \text{B}(40\hat{\text{i}}-8\hat{\text{j}})$ and $\text{C}(\text{a}\hat{\text{i}}+52\hat{\text{j}})$ are collinear, then a is equal to,
  • A
    40
  • ✓
    -40
  • C
    20
  • D
    -20
✓

Answer

Correct option: B.
-40
Given: Three points $\text{A}(60\hat{\text{i}}+3\hat{\text{j}}),\ \text{B}(40\hat{\text{i}}-8\hat{\text{j}})$ and $\text{C}(\text{a}\hat{\text{i}}+52\hat{\text{j}})$ are collinear. Then,

$\overrightarrow{\text{AB}}=\lambda\overrightarrow{\text{BC}}$

We have,

$\overrightarrow{\text{AB}}=\big(40\hat{\text{i}}-8\hat{\text{j}}\big)-\big(60\hat{\text{i}}+3\hat{\text{j}}\big)$

$=-20\hat{\text{i}}-11\hat{\text{j}}$

$\overrightarrow{\text{BC}}=\big(\text{a}\hat{\text{i}}-52\hat{\text{j}}\big)-\big(40\hat{\text{i}}-8\hat{\text{j}}\big)$

$=(\text{a}-40)\hat{\text{i}}-44\hat{\text{j}}$

Therefore,

$\overrightarrow{\text{AB}}=\lambda\overrightarrow{\text{BC}}$

$\Rightarrow-20\hat{\text{i}}-11\hat{\text{j}}=\lambda(\text{a}-40)\hat{\text{i}}-\lambda44\hat{\text{j}}$

$\Rightarrow\lambda(\text{a}-40)=-20,\ -44\lambda=-11\Rightarrow\lambda=\frac{1}4$

$\Rightarrow\text{a}-40=-80$

$\Rightarrow\text{a}=-40$

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