Question

f vector A = i cap - 2 j cap - 3 k cap , vector B= 2 i cap - j cap - k cap and Vector C = i cap + 3 j cap - 2 k cap find (vector A × vector B )× vector C​.


and the answer is I cap + j cap+ k cap



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Answer 1

Answer:

Let the vectors be A and B.

Given, A= i + 2j + 3k and B= 3i + 2j + 2k

We can find product of vectors in two ways

I. FINDING VECTOR/CROSS PRODUCT:

Using multiplication of vectors by matrix,

A × B= \begin{gathered}\left[\begin{array}{ccc}i&j&k\\1&2&3\\3&2&2\end{array}\right]\end{gathered}⎣⎢⎡i13j22k32⎦⎥⎤ = [(2x2)-(3x2)] i + [(1x2)-(3x3)] j + [(1x2)-(2x3)] k

A × B= [4-6] i + [2-9] j + [2-6] k =  –2 i – 7 j – 4 k

∴ Vector product of vectors A and B = -2i - 7 j - 4 k

II. FINDING SCALAR/DOT PRODUCT:

A · B = (1X3)i + (2X2)j + (3X2)k = 3i + 4j + 6k

∴ Scalar product of vectors A and B = 3i + 4j + 6k

Answer 2

Answer:

$\mathrm{\hat{\imath} + \hat{\jmath} +2\hat k}$

Step-by-step explanation:

Given :

$\mathrm{\overrightarrow{A} = \hat{\imath} -2 \hat{\jmath} -3 \hat k}$

$\mathrm{\overrightarrow{B} = 2\hat{\imath} - \hat{\jmath} - \hat k}$

$\mathrm{\overrightarrow C = \hat{\imath} +3 \hat{\jmath} -2 \hat k}$

To find :

$\mathrm{\Big(\overrightarrow{A} \times \overrightarrow{B}\Big) \times \overrightarrow{C}}$

Solution :

• 1st method :

We know that,

$\boxed{\boxed{\begin{array}{cc}\text{If \mathrm{\overrightarrow{A} = x\hat{\imath} + y\hat{\jmath} + z\hat k} and \mathrm{\overrightarrow{B} = l\hat{\imath} + m\hat{\jmath} + n\hat k} }\\\text{then } \vec A \times \vec B\ is \\\boxed{\vec A \times \vec B = \left|\begin{array}{ccc}i&j&k\\x&y&z\\l&m&n\end{array}\right| } \end{array}}}$

So, applying this

$\mathrm{\overrightarrow{A} \times \mathrm{\overrightarrow{B} = \left|\begin{array}{ccc}i&j&k\\1&-2&-3\\2&-1&-1\end{array}\right|}}$

$\implies \mathrm{\overrightarrow{A} \times \mathrm{\overrightarrow{B} = i(2-3)-j(-1+6) + k(-1+4)}}$

$\implies \mathrm{\overrightarrow{A} \times \overrightarrow{B} = -1 \hat{\imath}- 5\hat{\jmath} + 3\hat k}$

Now, finding $\mathrm{\Big(\overrightarrow{A} \times \overrightarrow{B}\Big) \times \overrightarrow{C}}$

Applying the same formula

$\mathrm{\Big(\overrightarrow{A} \times \overrightarrow{B}\Big) \times \overrightarrow{C} = \left|\begin{array}{ccc}i&j&k\\-1&-5&3\\1&3&-2\end{array}\right|}$

$\implies \mathrm{\Big(\overrightarrow{A} \times \overrightarrow{B}\Big) \times \overrightarrow{C} = i(10-9) - j(2 - 3) + k(-3 + 5)}$

$\implies \mathrm{\mathrm{\Big(\overrightarrow{A} \times \overrightarrow{B}\Big) \times \overrightarrow{C}} = \hat{\imath} + \hat{\jmath} + 2\hat k}$

__________________________

• 2nd Method :

We know that,

$\boxed{\boxed{\begin{array}{cc}\text{If \overrightarrow{A}, \overrightarrow{B} and \overrightarrow{C} are 3 vectors, then} \\\\ \mathrm{\Big(\overrightarrow{A} \times \overrightarrow{B}\Big) \times \overrightarrow{C} = \Big(\overrightarrow{A}.\overrightarrow{C}\Big)\overrightarrow{B} - \Big(\overrightarrow{B}.\overrightarrow{C}\Big)\overrightarrow{A}}\\\end{array}}}$

$\implies \mathrm{\overrightarrow{A}.\overrightarrow{C} = 1(1) -2(3)-3(-2)}$

$\implies \mathrm{\overrightarrow{A}.\overrightarrow{C} = 1}$

and

$\implies \mathrm{\overrightarrow{B}.\overrightarrow{C} = 2(1) -1(3)-1(-2)}$

$\implies \mathrm{\overrightarrow{B}.\overrightarrow{C} = 1}$

So,

$\implies \mathrm{\Big(\overrightarrow{A} \times \overrightarrow{B}\Big) \times \overrightarrow{C} = \Big(\overrightarrow{A}.\overrightarrow{C}\Big)\overrightarrow{B} - \Big(\overrightarrow{B}.\overrightarrow{C}\Big)\overrightarrow{A}}$

$\implies \mathrm{\Big(\overrightarrow{A} \times \overrightarrow{B}\Big) \times \overrightarrow{C} = (1)(\mathrm{2\hat{\imath} - \hat{\jmath} - \hat k}) - (1)(\mathrm{ \hat{\imath} -2 \hat{\jmath} -3 \hat k})}$

$\implies \mathrm{\Big(\overrightarrow{A} \times \overrightarrow{B}\Big) \times \overrightarrow{C} = \mathrm{2\hat{\imath} - \hat{\jmath} - \hat k} - \mathrm{ \hat{\imath} +2 \hat{\jmath} +3 \hat k}}$

$\implies \mathrm{\Big(\overrightarrow{A} \times \overrightarrow{B}\Big) \times \overrightarrow{C} = \mathrm{\hat{\imath} + \hat{\jmath} +2\hat k}}$

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From both the methods we are arriving at the same answer.

Hence, given answer is wrong.

The correct answer is :

$\boxed{\mathrm{\Big(\overrightarrow{A} \times \overrightarrow{B}\Big) \times \overrightarrow{C} = \mathrm{\hat{\imath} + \hat{\jmath} +2\hat k}}}$

Hope it helps!!