The diagonals of a parallelogram are the vectors 3 hati+6 hatj-2 hatk and - hati-2 hatj-8 hatk then the length of the shorter side of parallelogram is at.-

Question

The diagonals of a parallelogram are the vectors 3 hati+6 hatj-2 hatk and - hati-2 hatj-8 hatk then the length of the shorter side of parallelogram is at.- (A) 2 √3 (B) √14 (C) √29 (D) 4 √3

Tardigrade Admin · Accepted Answer

Q. The diagonals of a parallelogram are the vectors $3 \hat{i} + 6 \hat{j} - 2 \hat{k}$ and $- \hat{i} - 2 \hat{j} - 8 \hat{k}$ then the length of the shorter side of parallelogram is at.-

$23$

$14$

$29$

$43$

$\overline{a} = \frac{d _{1} + d _{2}}{2} = \frac{2 i + 4 j - 10 k}{2}$
$= i + 2 j - 5 k$
$∣ \overline{a} ∣ = 30$
$∣ \overline{b} ∣ = \frac{d _{1} - d _{2}}{2} = \frac{4 i + 8 j + 6 k}{2}$
$= 2 i + 4 j + 3 k$
$∣ \overline{b} ∣ = \frac{2}{4 + 16 + 9} = 29$

Q. The diagonals of a parallelogram are the vectors 3i^+6j^​−2k^ and −i^−2j^​−8k^ then the length of the shorter side of parallelogram is at.-

23​

14​

29​

43​

a=2d1​+d2​​=22i+4j−10k​ =i+2j−5k ∣a∣=30​ ∣b∣=2d1​−d2​​=24i+8j+6k​ =2i+4j+3k ∣b∣=4+16+9​2​=29​

Q. The diagonals of a parallelogram are the vectors $3 \hat{i} + 6 \hat{j} - 2 \hat{k}$ and $- \hat{i} - 2 \hat{j} - 8 \hat{k}$ then the length of the shorter side of parallelogram is at.-

$23$

$14$

$29$

$43$

$\overline{a} = \frac{d _{1} + d _{2}}{2} = \frac{2 i + 4 j - 10 k}{2}$
$= i + 2 j - 5 k$
$∣ \overline{a} ∣ = 30$
$∣ \overline{b} ∣ = \frac{d _{1} - d _{2}}{2} = \frac{4 i + 8 j + 6 k}{2}$
$= 2 i + 4 j + 3 k$
$∣ \overline{b} ∣ = \frac{2}{4 + 16 + 9} = 29$