The vectors from origin to the points A and B are veca = 2 hati -3 hatj + hatk and vecb = 2 hati +3 hatj + hatk , respectively, then the area of triangle OAB is

Question

The vectors from origin to the points A and B are veca = 2 hati -3 hatj + hatk and vecb = 2 hati +3 hatj + hatk , respectively, then the area of triangle OAB is (A) 340 (B) √25 (C) √229 (D) (1/2)√229

Tardigrade Admin · Accepted Answer

veca = 2 hati -3 hatj + hatk and vecb = 2 hati +3 hatj + hatk veca × vecb = | hati& hatj& hatk 2&-3&2 2&3&1| = -9 hati +2 hatj + 12 hatk Area of Δ OAB =(1/2)| veca × vecb| = (1/2) √81 + 4 + 144 = (1/2) √229

Q. The vectors from origin to the points $A$ and $B$ are $a = 2 \hat{i} - 3 \hat{j} + \hat{k}$ and $b = 2 \hat{i} + 3 \hat{j} + \hat{k}$ , respectively, then the area of triangle $O A B$ is

$340$

$25$

$229$

$\frac{1}{2} 229$

$a = 2 \hat{i} - 3 \hat{j} + \hat{k}$ and $b = 2 \hat{i} + 3 \hat{j} + \hat{k}$
$a \times b = ∣ ∣ \hat{i} 22 \hat{j} - 3 3 \hat{k} 21 ∣ ∣$
$= - 9 \hat{i} + 2 \hat{j} + 12 \hat{k}$
Area of $Δ O A B = \frac{1}{2} ∣ ∣ a \times b ∣ ∣$
$= \frac{1}{2} 81 + 4 + 144$
$= \frac{1}{2} 229$

Q. The vectors from origin to the points A and B are a=2i^−3j^​+k^ and b=2i^+3j^​+k^, respectively, then the area of triangle OAB is

340

25​

229​

21​229​