If a -2-3-k- b -k- c -2- - k- then area of parallelogram having diagonals a -b and b -c in sq- units is

Let x = a + nbsp; b = (2î 3ĵ+k̂)+( î+k̂) = î 3ĵ+2 k̂ Let y = b+ c = ( î+k̂)+(2ĵ k̂) = î+2ĵ Area of parallelogram = 12 | nbsp; x × nbsp; y | ⇒ 12 |( ...

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Standard VIII

Mathematics

Angles of a Parallelogram

If a =2î-3ĵ+...

Question

If $\to a = 2^i - 3^j +^k, \to b = -^i +^k, \to c = 2^j -^k,$ then area of parallelogram having diagonals $\to a + \to b$ and $\to b + \to c$ (in sq. units) is

\sqrt{21}

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\frac{\sqrt{21}}{2}

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\sqrt{19}

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\frac{\sqrt{19}}{2}

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Solution

The correct option is B $\frac{\sqrt{21}}{2}$
Let $\to x = \to a + \to b = (2^i - 3^j +^k) + (-^i +^k) =^i - 3^j + 2^k$
Let $\to y = \to b + \to c = (-^i +^k) + (2^j -^k) = -^i + 2^j$
Area of parallelogram $= \frac{1}{2} | \to x \times \to y |$
$\Rightarrow \frac{1}{2} | (^i - 3^j + 2^k) \times (-^i + 2^j) | = \frac{1}{2} ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 1 & - 3 & 2 - 1 & 2 & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$= \frac{1}{2} | ((0 - 4)^i - (0 + 2)^j + (2 - 3)^k) | = \frac{1}{2} | (- 4^i - 2^j -^k) | = \frac{\sqrt{16 + 4 + 1}}{2} = \frac{\sqrt{21}}{2}$ sq. units

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Angles of a Parallelogram

Standard VIII Mathematics

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If →a=2^i−3^j+^k,→b=−^i+^k,→c=2^j−^k, then area of parallelogram having diagonals →a+→b and →b+→c (in sq. units) is

If $\to a = 2^i - 3^j +^k, \to b = -^i +^k, \to c = 2^j -^k,$ then area of parallelogram having diagonals $\to a + \to b$ and $\to b + \to c$ (in sq. units) is