Vectors Ques 1
- If lines $x=a y+b, \quad z=c y+d$ and $x=a^{\prime} z+b^{\prime}$, $y=c^{\prime} z+d^{\prime}$ are perpendicular, then
(2019 Main, 9 Jan II)
(a) $a b^{\prime}+b c^{\prime}+1=0$
(b) $b b^{\prime}+c c^{\prime}+1=0$
(c) $a a^{\prime}+c+c^{\prime}=0$
(d) $c c^{\prime}+a+a^{\prime}=0$
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Answer:
Correct Answer: 1.(c)
Solution: (c) Let $1^{\text {st }}$ line is $x=a y+b, z=c y+d$.
$ \Rightarrow \frac{x-b}{a}=y, \frac{z-d}{c}=y \Rightarrow \frac{x-b}{a}=y=\frac{z-d}{c} $
The direction vector of this line is $\mathrm{b}_1=a \hat{\mathrm{i}}+\hat{\mathrm{j}}+c \hat{\mathrm{k}}$.
Let $2^{\text {nd }}$ line is $x=a^{\prime} z+b^{\prime}, y=c^{\prime} z+d^{\prime}$.
$ \Rightarrow \frac{x-b^{\prime}}{a^{\prime}}=z, \frac{y-d^{\prime}}{c}=z \Rightarrow \frac{x-b^{\prime}}{a^{\prime}}=\frac{y-d^{\prime}}{c^{\prime}}=z $
The direction vector of this line is $\mathrm{b}_2=\alpha^{\prime} \hat{\mathrm{i}}+c^{\prime} \hat{\mathrm{j}}+\hat{\mathrm{k}}$.
$\because$ The two lines are perpendicular, therefore, $\mathrm{b}_1 \cdot \mathrm{b}_2=0$.
$\Rightarrow(a \hat{\mathrm{i}}+\hat{\mathrm{j}}+c \hat{\mathrm{k}}) \cdot\left(a^{\prime} \hat{\mathrm{i}}+c \hat{\mathrm{j}}+\hat{\mathrm{k}}\right)=0$
$\Rightarrow a a^{\prime}+c+c=0 \quad \Rightarrow a a^{\prime}+c+c=0$
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