The area (in sq. units) of the parallelogram whose diagonals are along the vectors 8 hati - 6 hatj and 3 hati + 4 hatj - 12 hatk, is :

Question

The area (in sq. units) of the parallelogram whose diagonals are along the vectors 8 hati - 6 hatj and 3 hati + 4 hatj - 12 hatk, is : (A) 26 (B) 65 (C) 20 (D) 52

Tardigrade Admin · Accepted Answer

d1× d2 = |i&j&k 8&-6&0 3&4&-12| = 72 hati-(-96) hatj+50 hatk = 5178 + 7056 + 2500 |d1× d2| = √16900 = 130 A = (1/2)|d1× d2| = (1/2)×130 = 65

Q. The area (in sq. units) of the parallelogram whose diagonals are along the vectors $8 \hat{i} - 6 \hat{j}$ and $3 \hat{i} + 4 \hat{j} - 12 \hat{k}$ , is :

26

65

20

52

$d_{1} \times d_{2} = ∣ ∣ i 83 j - 6 4 k 0 - 12 ∣ ∣$
$= 72 \hat{i} - (- 96) \hat{j} + 50 \hat{k}$
$= 5178 + 7056 + 2500$
$∣ d_{1} \times d_{2} ∣ = 16900 = 130$
$A = \frac{1}{2} ∣ d_{1} \times d_{2} ∣ = \frac{1}{2} \times 130$
$= 65$

Q. The area (in sq. units) of the parallelogram whose diagonals are along the vectors 8i^−6j^​ and 3i^+4j^​−12k^, is :

26

65

20

52

d1​×d2​=∣∣​i83​j−64​k0−12​∣∣​ =72i^−(−96)j^​+50k^ =5178+7056+2500 ∣d1​×d2​∣=16900​=130 A=21​∣d1​×d2​∣=21​×130 =65

Q. The area (in sq. units) of the parallelogram whose diagonals are along the vectors $8 \hat{i} - 6 \hat{j}$ and $3 \hat{i} + 4 \hat{j} - 12 \hat{k}$ , is :

$d_{1} \times d_{2} = ∣ ∣ i 83 j - 6 4 k 0 - 12 ∣ ∣$
$= 72 \hat{i} - (- 96) \hat{j} + 50 \hat{k}$
$= 5178 + 7056 + 2500$
$∣ d_{1} \times d_{2} ∣ = 16900 = 130$
$A = \frac{1}{2} ∣ d_{1} \times d_{2} ∣ = \frac{1}{2} \times 130$
$= 65$