First slide

Cross product of vectors or vector product of vectors

Question

If the vertices of a triangle are $(1, 2, 3), (3, 4, - 1), (5, - 4, 2)$ and the area of triangle is $λ$ , then the number of positive integral divisors of $(\frac{λ^{2}}{6} + \sqrt[3]{λ^{2} + 25})$ is

Moderate

A

10
B

12
C

13
D

14

Solution

$T h e a r e a o f t h e t r i a n g l e f o r m e d b y t h e v e r t i c e s A, B, C i s \frac{1}{2} |\bar{A B} \times \bar{A C}|$
Here,
             $\begin{matrix} \bar{A B} = \bar{O B} - \bar{O A} \\ = 3 i + 4 j - k - i - 2 j - 3 k \\ = 2 i + 2 j - 4 k \\ A C = O C - O A \\ = 5 i - 4 j + 2 k - i - 2 j - 3 k \\ = 4 i - 6 j - k \end{matrix}$
Hence, the area of the triangle is
$\begin{matrix} \frac{1}{2} |\bar{A B} \times \bar{A C}| = \frac{1}{2} |\begin{matrix} i & j & k \\ 2 & 2 & - 4 \\ 4 & - 6 & - 1 \end{matrix}| \\ = \frac{1}{2} |i (- 2 - 24) - j (- 2 + 16) + k (- 12 - 8)| \\ = \frac{1}{2} |- 26 i - 14 j - 20 k| \\ = \sqrt{318} \end{matrix}$
  Consider ,
         $\begin{matrix} (\frac{λ^{2}}{6} + \sqrt[3]{λ^{2} + 25}) = \frac{318}{6} + \sqrt[3]{318 + 25} \\ = 53 + \sqrt[3]{343} \\ = 60 \\ = 2^{2} \cdot 3 \cdot 5 \end{matrix}$
The number of integral divisors of the above number is $(2 + 1) \cdot (1 + 1) \cdot (1 + 1) = 12$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App