Question

The diagonals of a parallelogram are represented by vector P vector = 5i cap - 4 j cap + 3 k cap and Q vector = 3 i cap + 2 cap - k cap .then the area of the parallelogram is

Answer 1

diagonals of parallelogram are ;

$d_1$ = 5i - 4j + 3k

$d_2$ = 3i + 2j - k

we know, area of parallelogram in terms of diagonals is given by,

$A=\frac{1}{2}|d_1||d_2|sin\alpha$

where $|d_1|$ and $|d_2|$ are the magnitude of diagonals of parallelogram and α is angle between them.

so, first find $|d_1|$ and $|d_2|$,

$|d_1|$ = √{5² + (-4)² + 3²} = √50

$|d_2|$ = √{3² + 2² + (-1)²} = √14

and angle between them, α = cos^-1$\left(\frac{d_1.d_2}{|d_1||d_2|}\right)$

= cos^-1{(5i -4j + 3k).(3i + 2j - k)}/√50.√14

= cos^-1{(15 - 8 - 3)/10√7}

= cos^-1(2/5√7)

so, sinα = sin(cos^-1(2/5√7)) = √171/5√7

so, area of parallelogram = 1/2 × √50 × √14 × √171/5√7

= 1/2 × 5√2 × (√7 × √2) × √171/5√7

= 1/2 × 2 × √171

= √171 sq unit .

Answer 2

ANSWER IS IN THE ATTACHMENT

I HOPE THIS WILL HELP YOU.