Math, asked by rajeevagam615, 6 months ago

w) Find a vector whose length is 3 and which is perpendicular to both the vectors a =
a = 3i+j - 4k
and b = 6i+4i-2k​

Answers

Answered by s115499ailma02691
0

Answer:

I can no slove yore question

Answered by aksharabininair
0

Step-by-step explanation:

A vector perpendicular to both A and B is obtained as the vector product of vectors A and B. Let C be that vector. That is,

C = A × B .

A = 3 i + j - 4 k and B = 6 i + 5 j - 2 k. Using Levi-Civita symbol € ijk, and unit vectors e

A × B = € ijk ei A j Bk . ( ijk, i,j, and k are all subscripts), where summation over the repeated symbol i is understood ( Einstein summation convention). The Levi-Civita symbol

€ijk = 0 if any two of the indices i,j,k, are same,

= +1 if all three indices i,j,k are distinct and are cylic in 1, 2, 3,

= - 1 if all three indices are distinct and are not cyclic in 1,2,3.

The indices i, j, k, take the values 1,2,3 . The unit vectors e i take the value e1 = i , e 2= j and e 3 = k. In this notation,

A ×B = i (€123 A² B³ - €132A³B²)+ j ( €231 A³ B¹ - €213A¹ B³) + k ( €312 A¹ B² - €321 A² B¹)= i [( 1×-2) -(-4×5)] + j [ (-4×6)-(3×-2)] + k [( 3×5)-(1×6)] = 18 i - 18 j + 9 k.

So A× B = 18 i - 18 j + 9 k.

| A×B | = [ (18)² + (-18)² +(9)² ]^½ = ( 324+324+81)^½= √729= 27.

So a vector perpendicular to both A and B is C, where

C = 18 i - 18 j + 9 k.

A unit vector along C is C/ 27. And a vector along C of magnitude 3 is 3 C/ 27 = C /9 = 2 i - 2 j + k.

So a vector of magnitude 3 and perpendicular to two given vectors A and B = 2 i -2 j + k..

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