Question

Find the area of parallelogram whose adjacent sides are given by the vectors a= 3i+j+4k and b= i-j+k.

Answer 1

Given, Parallelogram with adjacent sides

a = 3i + j + 4k

b = i - j + k

i, j, k are Unit vectors along x, y, z axes.

Area of the parallelogram is, $|a \times b|$

$a \times b \\ \\ = \begin{vmatrix}i & j & k\\3 & 1& 4 \\ 1& - 1& 1\end{vmatrix} \\ \\ = i(1 \times 1 - 4 \times - 1) - j(3 \times 1 - 4 \times 1) + k(3 \times - 1 - 1 \times 1) \\ \\ = i(1 + 4) - j( - 1) + k( - 3 - 1) \\ \\ = 5i + j - 4k</p><p>$

Now,

$|a \times b| = \sqrt{ {5}^{2} + {( 1)}^{2} + {( - 4)}^{2} } \\ \\ |a \times b| = \sqrt{25 + 1 + 16} \\ \\ |a \times b| = \sqrt{42}$

Therefore, Area of the parallelogram is √42 sq.units

Answer 2

Given, Parallelogram with adjacent sides

a = 3i + j + 4k

b = i - j + k

i, j, k are Unit vectors along x, y, z axes.

Area of the parallelogram is, |a × b|

$\begin{gathered}a \times b =\begin{vmatrix} i & j & k\\3 & 1& 4 \\ 1& - 1& 1\end{vmatrix} \\ \\ = i(1 \times 1 - 4 \times - 1) - j(3 \times 1 - 4 \times 1) + k(3 \times - 1 - 1 \times 1) \\ \\ = i(1 + 4) - j( - 1) + k( - 3 - 1) \\ \\ = 5i + j - 4k \end{gathered} </p><p>$

$=i(1×1−4×−1)−j(3×1−4×1)+k(3×−1−1×1) \\ =i(1+4)−j(−1)+k(−3−1) \\ =5i+j−4k$

Now,

$\begin{gathered} |a \times b| = \sqrt{ {5}^{2} + {( 1)}^{2} + {( - 4)}^{2} } \\ \\ |a \times b| = \sqrt{25 + 1 + 16} \\ \\ |a \times b| = \sqrt{42} \end{gathered}$

Therefore, Area of the parallelogram is √42 sq.units