Question

4. Find the unit vector perpendicular to both
vectors A=21- 2j+2k and B = 31- 2j+4K *
(4 Points)
-2i+j-2k /3.45
-2i+j-4k /2.45
x
-2i-j+k/2.45
0 -2i+3j+k /3.45​

Answer 1

Answer:

1/√6 [ -2i - j +k ]

Explanation:

unit vector : A × B / |A × B|

A × B = | ( i j k, 21- 2j+2k, 31- 2j+4K ) | a 3×3 matrix

A × B = i (-8+4) - j ( 8-6) + k ( -4+6)

A × B = -4i - 2j +2k

|A × B| = √ [ (-4)^2 + (-2)^2 + 2^2 ]

|A × B| = √24 = 2√6

unit vector = [ -4i - 2j +2k ] / 2√6

unit vector = 1/√6 [ -2i - j +k ]

Answer 2

Correct Question -

Find the unit vector perpendicular to both vector

$\vec{A} = 2\hat{\imath} + 2\hat{\jmath} + 2\hat{k}$

$\vec{B} = 3\hat{\imath} + 2\hat{\jmath} + 4\hat{k}$

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Solution

The cross product of two vectors yields a vector perpendicular to both the vector. So if$\vec{C}$ is perpendicular to $\vec{A}$and $\vec{B}$ then

$\vec { C } = \vec{A} \times \vec { B}$

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$\vec{A} = 2\hat{\imath} + 2\hat{\jmath} + 2\hat{k}$

$\longrightarrow$$A_1 = 2$

$\longrightarrow$$A_2 = 2$

$\longrightarrow$$A_3 = 2$

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$\vec{B} = 3\hat{\imath} + 2\hat{\jmath} + 4\hat{k}$

$\longrightarrow$$B_1 = 3$

$\longrightarrow$$B_2 = 2$

$\longrightarrow$$B_3 = 4$

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$\vec { C } = ( A_2B_3 - B_2A_3 ) \hat{\imath} - ( A_1B_3 - B_1A_3 ) \hat{\jmath} + ( A_1B_2 - B_1A_2) \hat{k}$

Putting the values -

$\vec { C } = (2 \times 4 - 2 \times 2) \hat{\imath}-( 2 \times 4 - 3 \times 2) \hat{\jmath}+ (2 \times 2 - 3 \times 2)\hat{k}$

$\longrightarrow$$= 4\hat{\imath} - 2 \hat{\jmath}- 2\hat{k}$

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Unit vector in direction of C =$\frac{\vec{C}}{ |\vec{C}| }$

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$\longrightarrow$$|\vec{C}| = \sqrt{ {4}^{2} + {2}^{2} + 2 {}^{2} } = \sqrt{24}$

$\longrightarrow$ $|\vec{C}| = 2\sqrt{6}$

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Unit vector =$\frac{4\hat{\imath} - 2 \hat{\jmath}- 2\hat{k}}{2 \sqrt{6} }$

$\longrightarrow$= $\frac{2\hat{\imath} - \hat{\jmath}- \hat{k}}{ 2.45 }$

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