Question

find the area of parallelogram with adjusent sides form by vector A=2i-j+k and vector B=I+2j-k​

Answer 1

Answer:

Area of parallelogram formed by two adjacent sides in terms of vector is given by the formula:

⇒ Area = Magnitude or Modulus of Cross Product of Two Sides

Therefore the area of parallelogram in our case is:

⇒ Area = |A × B|

According to the question,

⇒ A = 2i - j + k

⇒ B = i + 2j - k

Calculating the Cross Product we get:

$\vec{A} \times \vec{B} = \left[\begin{array}{ccc}i&j&k\\2&-1&1\\1&2&-1\end{array}\right]\\\\\\\implies \vec{A} \times \vec{B} = i(1 - 2) - j(-2-1) + k(4+1)\\\\\implies \vec{A} \times \vec{B} = i(-1) -j(-3) + k(5)\\\\\implies \boxed{\bf{\vec{A} \times \vec{B} = -i + 3j + 5k}}$

Hence the magnitude or Modulus of  A × B is given as:

$\text{Area} = \sqrt{ (-1)^2 + (3)^2 + (5)^2}\\\\\text{Area} = \sqrt{1 + 9 + 25}\\\\\boxed{ \bf{\textbf{Area} = \sqrt{35}}}$

Hence the Area of the parallelogram is √35 units.

Answer 2

Answer:

Area of parallelogram formed by two adjacent sides in terms of vector is given by the formula:

⇒ Area = Magnitude or Modulus of Cross Product of Two Sides

Therefore the area of parallelogram in our case is:

⇒ Area = |A × B|

According to the question,

⇒ A = 2i - j + k

⇒ B = i + 2j - k

$</p><p>\tt{\implies}{ \boxed{\mathbb{\pink{alculating the Cross Product we get:}}}} \\ \\ </p><p></p><p>\tt{\implies}e{ {\mathbb{\blue{\begin{gathered}\vec{A} \times \vec{B} = \left[\begin{array}{ccc}i&j&k\\2&-1&1\\1&2&-1\end{array}\right] \\ \\\\ \\ \\\implies \vec{A} \times \vec{B} = i(1 - 2) - j(-2-1) + k(4+1)\ \\ \\\tt \: \implies \vec{A} \times \vec{B} = i(-1) -j(-3) + k(5)\\\\ \\ \ \boxed{\bf{\vec{A} \times \vec{B} = -i + 3j + 5k}}\end{gathered} }}}}</p><p>A</p><p> × </p><p>B</p><p> = </p><p>⎣</p><p>⎢</p><p>⎡</p><p></p><p> </p><p>i</p><p>2</p><p>1</p><p></p><p> </p><p>j</p><p>−1</p><p>2</p><p></p><p> </p><p>k</p><p>1</p><p>−1</p><p></p><p> </p><p>⎦</p><p>⎥</p><p>⎤</p><p>$[/tex]

$\tt{\implies}{ \boxed{\mathbb{\green{Hence \: the \: magnitude \: or Modulus \: of \:  A × B \: is given as:}}}} \\ </p><p></p><p>\begin{gathered}\text{Area} = \sqrt{ (-1)^2 + (3)^2 + (5)^2}\\\\\text{Area} = \sqrt{1 + 9 + 25} \\ \\\\\boxed{ \bf{\textbf{Area} = \sqrt{35}}}\end{gathered}\tt{\implies}{ \boxed{\mathbb{\orange{Area=(−1)2+(3)2+(5)2 }}}}\\ Area=1+9+25Area=35 \\ </p><p></p><p>\tt{\implies}{ \boxed{\mathbb{\red{Hence the Area of the parallelogram is √35 units}}}}.$

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