Question

Find the area of parallelogram determined by the vectors i +2j+3k and 3i-2j+k

Answer 1

Answer:

Area of the parallelogram is $8\sqrt{3}$ square units

Step-by-step explanation:

Formula used:

Area of parallelogram having adjacent sides $\vec{a}\:and\:\vec{b}$ is $|\vec{a}\:X\:\vec{b}|$

Let

$\vec{a}=\vec{i}+2\vec{j}+3\vec{k}$

$\vec{b}=3\vec{i}-2\vec{j}+\vec{k}$

$\vec{a}\:X\:\vec{b}=\left|\begin{array}{ccc}\vec{i}&\vec{j}&\vec{k}\\1&2&3\\3&-2&1\end{array}\right|$

$\vec{a}\:X\:\vec{b}=\vec{i}(2+6)-\vec{j}(1-9)+\vec{k}(-2-6)$

$\vec{a}\:X\:\vec{b}=8\vec{i}+8\vec{j}-8\vec{k}$

$\vec{a}\:X\:\vec{b}=8(\vec{i}+\vec{j}-\vec{k})$

$|\vec{a}\:X\:\vec{b}|=8\sqrt{1^2+1^2+(-1)^}$

$|\vec{a}\:X\:\vec{b}|=8\sqrt{1^2+1^2+(-1)^}$

$|\vec{a}\:X\:\vec{b}|=8\sqrt{3}$

Area of the parallelogram is $8\sqrt{3}$ square units

Answer 2

Step-by-step explanation:

Formula used:

Area of parallelogram having adjacent sides \vec{a}\:and\:\vec{b}

a

and

b

is |\vec{a}\:X\:\vec{b}|∣

a

X

b

∣

Let

\vec{a}=\vec{i}+2\vec{j}+3\vec{k}

a

=

i

+2

j

+3

k

\vec{b}=3\vec{i}-2\vec{j}+\vec{k}

b

=3

i

−2

j

+

k

\begin{gathered}\vec{a}\:X\:\vec{b}=\left|\begin{array}{ccc}\vec{i}&\vec{j}&\vec{k}\\1&2&3\\3&-2&1\end{array}\right|\end{gathered}

a

X

b

=

∣

∣

∣

∣

∣

∣

∣

i

1

3

j

2

−2

k

3

1

∣

∣

∣

∣

∣

∣

∣

\vec{a}\:X\:\vec{b}=\vec{i}(2+6)-\vec{j}(1-9)+\vec{k}(-2-6)

a

X

b

=

i

(2+6)−

j

(1−9)+

k

(−2−6)

\vec{a}\:X\:\vec{b}=8\vec{i}+8\vec{j}-8\vec{k}

a

X

b

=8

i

+8

j

−8

k

\vec{a}\:X\:\vec{b}=8(\vec{i}+\vec{j}-\vec{k})

a

X

b

=8(

i

+

j

−

k

)

|\vec{a}\:X\:\vec{b}|=8\sqrt{1^2+1^2+(-1)^}

|\vec{a}\:X\:\vec{b}|=8\sqrt{1^2+1^2+(-1)^}

|\vec{a}\:X\:\vec{b}|=8\sqrt{3}∣

a

X

b

∣=8

3

Area of the parallelogram is 8\sqrt{3}8

3

square units