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Length of Subtangent

If a = 2 i ...

Question

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

A

\sqrt{21}

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B

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

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C

\sqrt{19}

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D

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

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Solution

The correct option is B $\frac{\sqrt{21}}{2}$
$\begin{matrix} ¯ ¯ ¯ a = 2 ˆ i - 3 ˆ j + ˆ k, ¯ ¯ b = - ˆ i + ˆ k ¯ ¯ c = 2 ˆ j - ˆ k s o, d i a g o n a l s (¯ ¯ ¯ a + ¯ ¯ b) = (ˆ i - 3 ˆ j + 2 ˆ k) a n d (¯ ¯ b + ¯ ¯ c) = (- ˆ i + 2 ˆ j) A r e a o f p a r a l l e log r a m = \frac{1}{2} | d_{1} \times d_{2} | = \frac{1}{2} \times \sqrt{21} = \frac{\sqrt{21}}{2} O p t i o n B i s c o r r e c t a n s w e r . \end{matrix}$

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Similar questions

a = 2 i - 3 j + k, b = - i + k, c = 2 j - k,

find the area of the parallelogram having diagonals a+band b+c.

\vec{a} = 2 \hat{i} - 3 \hat{j} + \hat{k}, \vec{b} = - \hat{i} + \hat{k}, \vec{c} = 2 \hat{j} - \hat{k}

are three vectors, find the area of the parallelogram having diagonals

(\vec{a} + \vec{b})

(\vec{b} + \vec{c})

[\vec{a} \vec{b} \vec{c}]

\vec{a} = 2 \hat{i} - 3 \hat{j}, \vec{b} = \hat{i} + \hat{j} - \hat{k} and \vec{c} = 3 \hat{i} - \hat{k}

\vec{a} = \hat{i} - 2 \hat{j} + 3 \hat{k}, \vec{b} = 2 \hat{i} + \hat{j} - \hat{k} and \vec{c} = \hat{j} + \hat{k}

\vec{a} = 2 \hat{i} + 3 \hat{j} + \hat{k}, \vec{b} = \hat{i} - 2 \hat{j} + \hat{k} and \vec{c} = - 3 \hat{i} + \hat{j} + 2 \hat{k}

a = i + j + k, b = i + 3 j + 5 k

c = 7 i + 9 j + 11 k

, then the area of Parallelogram having diagonals a+b and b+c is.

\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

\to b + \to c

(in sq. units) is

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