First slide

Algebra of vectors

Question

A, B, C and D are four points such that $\vec{AB} = m (2 \hat{i} - 6 \hat{j} + 2 \hat{k}), \vec{BC} = (\hat{i} - 2 \hat{j})$ and $\vec{CD} = n (- 6 \hat{i} + 15 \hat{j} - 3 \hat{k})$ If CD intersects AB at some point E, then

Moderate

A

$m \geq 1 / 2$
B

$n \geq 1 / 3$
C

m=n
D

m<n

Solution

$Let \vec{EB} = p \vec{AB} and \vec{CE} = q \vec{CD}$
$Then 0 < p and q \leq 1$
$\begin{array}{l} since \vec{EB} + \vec{BC} + \vec{CE} = \vec{0} \\ pm (2 \hat{i} - 6 \hat{j} + 2 \hat{k}) + (\hat{i} - 2 \hat{j}) + qn (- 6 \hat{i} + 15 \hat{j} - 3 \hat{k}) = \vec{0} \end{array}$
$or (2 pm + 1 - 6 qn) \hat{i} + (- 6 pm - 2 + 15 qn) \hat{j} + (2 pm - 6 qn) \hat{k} = \vec{0}$
$\Rightarrow 2 pm - 6 qn + 1 = \vec{0}, - 6 pm - 2 + 15 qn = \vec{0}, 2 pm - 6 qn = \vec{0}$
Solving these, we get
$\begin{array}{l} p = 1 / (2 m) and q = 1 / (3 n) \\ ∴ 0 < 1 / (2 m) \leq 1 and 0 < 1 / (3 n) \leq 1 \\ \Rightarrow m \geq 1 / 2 and n \geq 1 / 3 \end{array}$

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