$A, B, C$ and $D$ are four points such that $A B = m (2 i - 6 j + 2 k), B C = i - 2 j$ and $C D = n (- 6 i + 15 j - 3 k)$ . If $A B$ and $C D$ intersect at some point $E$ , then

m \geq \frac{1}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

m \geq \frac{1}{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Area of

△ B C E = \frac{1}{2 m} \sqrt{6}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

all of these

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct options are
A $m \geq \frac{1}{2}$
B $m \geq \frac{1}{3}$
C all of these
D Area of $△ B C E = \frac{1}{2 m} \sqrt{6}$
Let $E B = p A B, C E = q C D$
then, $0 < p, q \leq 1$
Since $\to E B + \to B C + \to C E = 0$
$∴ p m (2 i - 6 j + 2 k) + (i - 2 j) + q n (- 6 i + 15 j - 3 k) = 0$
$\Rightarrow (2 p m + 1 - 6 q n) i + (- 6 p m - 2 + 15 q n) j + (2 p m - 6 q n) k = 0$
$\Rightarrow 2 p m - 6 q n + 1 = 0, - 6 p m - 2 + 15 q n = 0 \Rightarrow 2 p m - 6 q n = 0$
Solving these, we get
$p = \frac{1}{(2 m)}$ and $q = \frac{1}{(3 n) .}$
$∴ 0 < \frac{1}{(2 m)} \leq$ and $0 < \frac{1}{(3 n)} \leq 1$
$\Rightarrow m \geq \frac{1}{2}$ and $n \geq \frac{1}{3.}$
The area of $△ B C E = \frac{1}{2} | E B \times B C | = \frac{1}{2 m} \sqrt{6.}$

Suggest Corrections

Similar questions

A, B, C and D are four points such that $- - \to A B = m (2^i - 6^j + 2^k), - - \to B C = (^i - 2^j) and - - \to C D = n (- 6^i + 15^j - 3^k) .$ If $^C D$ intersects $- - \to A B$ at some point E, then which of the following option(s) is/are correct?