First slide

Algebra of vectors

Question

$T h r e e l i n e s$ $L_{1} : \vec{r} = λ \hat{i}, λ \in ℝ$ , $L_{2} : \vec{r} = \hat{k} + μ \hat{j}, μ \in ℝ$ , $L_{3} : \vec{r} = \hat{i} + \hat{j} + ν \hat{k}, ν \in ℝ$ $a r e g i v e n . F o r w h i c h t h e p o i n t s$ $Q$ $o n$ $L_{2}$ $c a n w e f i n d$ $P$ $o n$ $L_{1}$ $a n d a p o i n t R o n L_{3} s o t h a t P, Q, R a r e c o l l i n e a r$

Difficult

A

$\hat{k} - \frac{1}{2} \hat{j}$
B

$\hat{k} + \frac{1}{2} \hat{j}$
C

$\hat{k}$
D

$\hat{k} + \hat{j}$

Solution

$Position vector of point P i s \vec{p} = λ \hat{i}$
$Position vector of Q i s \hat{q} = μ \hat{j} + \hat{k}$
$P o s i t i o n v e c t o r o f point R i s \vec{r} = \hat{i} + \hat{j} + r \hat{k}$
$PQR are collinear. Hence x (\vec{PQ}) = y (\vec{PR})$
$\Rightarrow x (- λ i + μ j + k) = y ((1 - λ) i + j + r k) I t g i v e s \frac{x}{y} = \frac{1 - λ}{- λ} = \frac{1}{μ} = r$
$\Rightarrow \vec{q} = \frac{1}{r} \hat{j} + \hat{k} or \vec{q} = \frac{λ}{λ - 1} \hat{j} + \hat{k}, where r \neq 0, λ \neq 0, \frac{λ}{λ - 1} \neq 1$
$\Rightarrow μ \neq 0, 1$
$Hence, \vec{q} \neq \hat{k} or \hat{j} + \hat{k}$

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