JEE MAIN - Mathematics (2019 - 10th January Evening Slot - No. 10)
If $$\overrightarrow \alpha $$ = $$\left( {\lambda - 2} \right)\overrightarrow a + \overrightarrow b $$ and $$\overrightarrow \beta = \left( {4\lambda - 2} \right)\overrightarrow a + 3\overrightarrow b $$ be two given vectors $$\overrightarrow a $$ and $$\overrightarrow b $$ are non-collinear. The value of $$\lambda $$ for which vectors $$\overrightarrow \alpha $$ and $$\overrightarrow \beta $$ are collinear, is -
4
3
$$-$$3
$$-$$4
Explanation
$$\overrightarrow \alpha = \left( {\lambda - 2} \right)\overrightarrow \alpha + \overrightarrow b $$
$$\overrightarrow \beta = \left( {4\lambda - 2} \right)\overrightarrow \alpha + 3\overrightarrow b $$
$${{\lambda - 2} \over {4\lambda - 2}} = {1 \over 3}$$
$$3\lambda - 6 = 4\lambda - 2$$
$$\lambda = - 4$$
$$\overrightarrow \beta = \left( {4\lambda - 2} \right)\overrightarrow \alpha + 3\overrightarrow b $$
$${{\lambda - 2} \over {4\lambda - 2}} = {1 \over 3}$$
$$3\lambda - 6 = 4\lambda - 2$$
$$\lambda = - 4$$
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