
If $ a = 2i - 3j + k,\;{\text{b = - i + k, c = 2j - k,}} $ find the area of the parallelogram having diagonals $ a + b $ and $ b + c $
A.Area $ = \dfrac{1}{4}\sqrt {21} $
B.Area $ = \dfrac{1}{4}\sqrt {19} $
C.Area $ = \dfrac{1}{2}\sqrt {19} $
D.Area $ = \dfrac{1}{2}\sqrt {21} $
Answer
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Hint: First of all we will assume two diagonals then will find the diagonals then will place the value of diagonals in the area formula. Will use a cross-product method to find the diagonal product and then will find magnitude.
Complete step-by-step answer:
Let us take the given vertices –
$ a = 2i - 3j + k,\;{\text{b = - i + k, c = 2j - k,}} $
Let us assume that diagonals $ {d_1} = a + b $ and $ {d_2} = b + c $
Now, given that the diagonals of the parallelogram are –
Find $ a + b $
Place the values in the above expression –
$ a + b = (2i - 3j + k) + ( - i + k) $
Open the brackets and simplify. Remember when there is a positive sign outside the bracket then there is no change in the signs of the terms inside the bracket when you open it.
$ a + b = 2i - 3j + k - i + k $
Make the pair of like terms –
$ \Rightarrow a + b = \underline {2i - i} - 3j\underline { + k + k} $
Simplify the above equation –
$ \Rightarrow a + b = i - 3j + 2k $
$ \Rightarrow {d_2} = i - 3j + 2k $ .... (A)
Similarly for second diagonal
$ b + c = - i + k + 2j - k $
Make the pair of like terms –
$ b + c = - i + 2j\underline { - k + k} $
Terms with the same value and opposite sign cancel each other. Simplify the above equation –
$ b + c = - i + 2j $
Therefore, $ {d_2} = - i + 2j $ .... (B)
Now, the area of the parallelogram is
$ A = \dfrac{1}{2}\left| {{d_1} \times {d_2}} \right| $
Place values form equation (A) and (B)
$ A = \dfrac{1}{2}\left| {(i - 3j + 2k) \times ( - i + 2j)} \right| $ ..... (C)
Now find cross product –
$ \left| {\begin{array}{*{20}{c}}
i&j&k \\
1&{ - 3}&2 \\
{ - 1}&2&0
\end{array}} \right| $
Open the determinant –
$ = i( - 4) - j(2) + k(2 - 3) $
Simplify the above equation – when you subtract a bigger number from smaller there will be a negative sign in resultant value.
$ $ $ = 4i - 2j - k $
Now find the mode of the above vector expression –
$ \left| {4i - 2j - k} \right| = \sqrt {{4^2} + {{( - 2)}^2} + {{( - 1)}^2}} $
Remember the square of the negative number also gives the positive value. Since minus multiplied with minus gives plus.
$ \left| {4i - 2j - k} \right| = \sqrt {16 + 4 + 1} $
Simplify the above equation –
$ \left| {4i - 2j - k} \right| = \sqrt {21} $ .... (D)
Place above value in equation (C)
$ \Rightarrow A = \dfrac{{\sqrt {21} }}{2} $ Sq.units
So, the correct answer is “Option D”.
Note: Parallelogram law:
If two vectors are represented by two adjacent sides of a parallelogram, then the diagonal of parallelogram through the common point represents the sum of the two vectors in both magnitude and direction.
Complete step-by-step answer:
Let us take the given vertices –
$ a = 2i - 3j + k,\;{\text{b = - i + k, c = 2j - k,}} $
Let us assume that diagonals $ {d_1} = a + b $ and $ {d_2} = b + c $
Now, given that the diagonals of the parallelogram are –
Find $ a + b $
Place the values in the above expression –
$ a + b = (2i - 3j + k) + ( - i + k) $
Open the brackets and simplify. Remember when there is a positive sign outside the bracket then there is no change in the signs of the terms inside the bracket when you open it.
$ a + b = 2i - 3j + k - i + k $
Make the pair of like terms –
$ \Rightarrow a + b = \underline {2i - i} - 3j\underline { + k + k} $
Simplify the above equation –
$ \Rightarrow a + b = i - 3j + 2k $
$ \Rightarrow {d_2} = i - 3j + 2k $ .... (A)
Similarly for second diagonal
$ b + c = - i + k + 2j - k $
Make the pair of like terms –
$ b + c = - i + 2j\underline { - k + k} $
Terms with the same value and opposite sign cancel each other. Simplify the above equation –
$ b + c = - i + 2j $
Therefore, $ {d_2} = - i + 2j $ .... (B)
Now, the area of the parallelogram is
$ A = \dfrac{1}{2}\left| {{d_1} \times {d_2}} \right| $
Place values form equation (A) and (B)
$ A = \dfrac{1}{2}\left| {(i - 3j + 2k) \times ( - i + 2j)} \right| $ ..... (C)
Now find cross product –
$ \left| {\begin{array}{*{20}{c}}
i&j&k \\
1&{ - 3}&2 \\
{ - 1}&2&0
\end{array}} \right| $
Open the determinant –
$ = i( - 4) - j(2) + k(2 - 3) $
Simplify the above equation – when you subtract a bigger number from smaller there will be a negative sign in resultant value.
$ $ $ = 4i - 2j - k $
Now find the mode of the above vector expression –
$ \left| {4i - 2j - k} \right| = \sqrt {{4^2} + {{( - 2)}^2} + {{( - 1)}^2}} $
Remember the square of the negative number also gives the positive value. Since minus multiplied with minus gives plus.
$ \left| {4i - 2j - k} \right| = \sqrt {16 + 4 + 1} $
Simplify the above equation –
$ \left| {4i - 2j - k} \right| = \sqrt {21} $ .... (D)
Place above value in equation (C)
$ \Rightarrow A = \dfrac{{\sqrt {21} }}{2} $ Sq.units
So, the correct answer is “Option D”.
Note: Parallelogram law:
If two vectors are represented by two adjacent sides of a parallelogram, then the diagonal of parallelogram through the common point represents the sum of the two vectors in both magnitude and direction.
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