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3-1.Vectors
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જો સદિશ $ 2\hat i + 3\hat j - \hat k $ અને $ - 4\hat i - 6\hat j + \lambda \hat k $ સમાંતર હોય,તો $\lambda = $_______
A$0$
B$2$
C$3$
D$4$
Solution
(b) Let $\overrightarrow A = 2\hat i + 3\hat j – \hat k$ and $\overrightarrow B = – 4\hat i – 6\hat j + \lambda \hat k$
$\overrightarrow A $ and $\overrightarrow B $ are parallel to each other
$\frac{{{a_1}}}{{{b_1}}} = \frac{{{a_2}}}{{{b_2}}} = \frac{{{a_3}}}{{{b_3}}}$ i.e. $\frac{2}{{ – 4}} = \frac{3}{{ – 6}} = \frac{{ – 1}}{\lambda }$$ \Rightarrow \,\,\lambda = 2$.
$\overrightarrow A $ and $\overrightarrow B $ are parallel to each other
$\frac{{{a_1}}}{{{b_1}}} = \frac{{{a_2}}}{{{b_2}}} = \frac{{{a_3}}}{{{b_3}}}$ i.e. $\frac{2}{{ – 4}} = \frac{3}{{ – 6}} = \frac{{ – 1}}{\lambda }$$ \Rightarrow \,\,\lambda = 2$.
Standard 11
Physics
