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3-1.Vectors
hard
Consider three vectors $A =\hat{ i }+\hat{ j }-2 \hat{ k }, B =\hat{ i }-\hat{ j }+\hat{ k }$ and $C =2 \hat{ i }-3 \hat{ j }+4 \hat{ k }$. A vector $X$ of the form $\alpha A +\beta B$ ( $\alpha$ and $\beta$ are numbers) is perpendicular to $C$.The ratio of $\alpha$ and $\beta$ is
A$1: 1$
B$2: 1$
C$-1: 1$
D$3: 1$
Solution
(a)
The given vector, $X =\alpha A +\beta B =\alpha(\hat{ i }+\hat{ j }-2 \hat{ k })+\beta(\hat{ i }-\hat{ j }+\hat{ k })$
$=(\alpha+\beta) \hat{ i }+(\alpha-\beta) \hat{ j }+(-2 \alpha+\beta) \hat{ k }$
and $C =2 \hat{ i }-3 \hat{ j }+4 \hat{ k }$
Now, $\alpha A +\beta B$ is perpendicular to $C$ when
$(\alpha A +\beta B ) \cdot C =0$
$\Rightarrow(\alpha+\beta)(2)+(\alpha-\beta)(-3)+(-2 \alpha+\beta)(4)=0$
$\Rightarrow \quad 2 \alpha+2 \beta-3 \alpha+3 \beta-8 \alpha+4 \beta=0$
$\Rightarrow \quad-9 \alpha+9 \beta=0$
$\Rightarrow \quad-9 \alpha=-9 \beta \Rightarrow \frac{\alpha}{\beta}=1$
$\therefore \quad \alpha: \beta=1: 1$
The given vector, $X =\alpha A +\beta B =\alpha(\hat{ i }+\hat{ j }-2 \hat{ k })+\beta(\hat{ i }-\hat{ j }+\hat{ k })$
$=(\alpha+\beta) \hat{ i }+(\alpha-\beta) \hat{ j }+(-2 \alpha+\beta) \hat{ k }$
and $C =2 \hat{ i }-3 \hat{ j }+4 \hat{ k }$
Now, $\alpha A +\beta B$ is perpendicular to $C$ when
$(\alpha A +\beta B ) \cdot C =0$
$\Rightarrow(\alpha+\beta)(2)+(\alpha-\beta)(-3)+(-2 \alpha+\beta)(4)=0$
$\Rightarrow \quad 2 \alpha+2 \beta-3 \alpha+3 \beta-8 \alpha+4 \beta=0$
$\Rightarrow \quad-9 \alpha+9 \beta=0$
$\Rightarrow \quad-9 \alpha=-9 \beta \Rightarrow \frac{\alpha}{\beta}=1$
$\therefore \quad \alpha: \beta=1: 1$
Standard 11
Physics
