Find a → × b → and a → × b → if a → = 2 i ^ + j ^ + 3 k ^ and b → = 3 i ^ + 5 j ^ - 2 k ^

Given: a → = 2 i ^ + j ^ + 3 k ^ and b → = 3 i ^ + 5 j ^ - 2 k ^ Now we know, a → × b → = i ^ j ^ k ^ 2 1 3 3 5 - 2 ⇒ a → × b → = - 2 - 15 i ^ - - 4 - 9 j ^ + 10 - 3 k ^ ⇒ a → × b → = - 17 i ^ + 13 j ^ + 7 k ^ And the magnitude of a → × b → can be calculated using the following formula, a → × b → = - 17 2 + 13 2 + 7 2 ⇒ a → × b → = 289 + 169 + 49 ⇒ a → × b → = 507 .

Find a → × b → and a → × b → if a → = 2 i ^ + j ^ + 3 k ^ and b → = 3 i ^ + 5 j ^ - 2 k ^

Given: a → = 2 i ^ + j ^ + 3 k ^ and b → = 3 i ^ + 5 j ^ - 2 k ^ Now we know, a → × b → = i ^ j ^ k ^ 2 1 3 3 5 - 2 ⇒ a → × b → = - 2 - 15 i ^ - - 4 - 9 j ^ + 10 - 3 k ^ ⇒ a → × b → = - 17 i ^ + 13 j ^ + 7 k ^ And the magnitude of a → × b → can be calculated using the following formula, a → × b → = - 17 2 + 13 2 + 7 2 ⇒ a → × b → = 289 + 169 + 49 ⇒ a → × b → = 507 .

Find a → × b → and a → × b → if a → = i ^ - 7 j ^ + 7 k ^ and b → = 3 i ^ - 2 j ^ + 2 k ^

Given: a → = i ^ - 7 j ^ + 7 k ^ and b → = 3 i ^ - 2 j ^ + 2 k ^ Now we know, a → × b → = i ^ j ^ k ^ 1 - 7 7 3 - 2 2 ⇒ a → × b → = - 14 + 14 i ^ - 2 - 21 j ^ + - 2 + 21 k ^ ⇒ a → × b → = 0 i ^ + 19 j ^ + 19 k ^ ⇒ a → × b → = 19 j ^ + 19 k ^ . And the magnitude of a → × b → can be calculated using the following formula, a → × b → = 0 2 + 19 2 + 19 2 ⇒ a → × b → = 2 × 19 2 ⇒ a → × b → = 19 2 .

If a → = 2 i ^ - j ^ + k ^ and b → = 3 i ^ + 4 j ^ - k ^ , prove that a → × b → represents a vector which is perpendicular to both a → and b → .

Given: a → = 2 i ^ - j ^ + k ^ and b → = 3 i ^ + 4 j ^ - k ^ Now we know, a → × b → = i ^ j ^ k ^ 2 - 1 1 3 4 - 1 ⇒ a → × b → = 1 - 4 i ^ - - 2 - 3 j ^ + 8 + 3 k ^ ⇒ a → × b → = - 3 i ^ + 5 j ^ + 11 k ^ Now, a → × b → · a → = ( - 3 i ^ + 5 j ^ + 11 k ^ ) · ( 2 i ^ - j ^ + k ^ ) ⇒ a → × b → · a → = - 6 - 5 + 11 ⇒ a → × b → · a → = 0 Hence a → × b → ⊥ a → Similarly, a → × b → · b → = ( - 3 i ^ + 5 j ^ + 11 k ^ ) · ( 3 i ^ + 4 j ^ - k ^ ) ⇒ a → × b → · b → = - 9 + 20 - 11 ⇒ a → × b → · b → = 0 So, a → × b → ⊥ b → Therefore, a → × b → represents a vector which is perpendicular to both a → and b → .

If a → = 7 i ^ + 3 j ^ - 5 k ^ , b → = 2 i ^ + 5 j ^ - k ^ and c → = - i ^ + 2 j ^ + 4 k ^ . Find a → - b → × c → - b →

Given: a → = 7 i ^ + 3 j ^ - 5 k ^ , b → = 2 i ^ + 5 j ^ - k ^ and c → = - i ^ + 2 j ^ + 4 k ^ . First, we will find: a → - b → a → - b → = 7 i ^ + 3 j ^ - 5 k ^ - 2 i ^ + 5 j ^ - k ^ , ⇒ a → - b → = 5 i ^ - 2 j ^ - 4 k ^ . . . 1 Now we will find: c → - b → c → - b → = - i ^ + 2 j ^ + 4 k ^ - 2 i ^ + 5 j ^ - k ^ , ⇒ c → - b → = - 3 i ^ - 3 j ^ + 5 k ^ . . . 2 Now we will find: a → - b → × c → - b → We will substitute values from Equations 1 and 2 . Hence a → - b → × c → - b → = i ^ j ^ k ^ 5 - 2 - 4 - 3 - 3 5 , ⇒ a → - b → × c → - b → = - 10 - 12 i ^ - 25 - 12 j ^ + - 15 - 6 k ^ , ⇒ a → - b → × c → - b → = - 22 i ^ - 13 j ^ - 21 k . ^

Find a unit vector perpendicular to the plane of two vectors a → and b → , where a → = i ^ - j ^ and b → = j ^ + k ^

Given, a → = i ^ - j ^ and b → = j ^ + k ^ Now the cross product of these vectors will be a → × b → = i ^ j ^ k ^ 1 - 1 0 0 1 1 ⇒ a → × b → = - 1 - 0 i ^ - 1 - 0 j ^ + 1 - 0 k ^ ⇒ a → × b → = - i ^ - j ^ + k ^ And the corresponding magnitude will be a → × b → = - 1 2 + - 1 2 + 1 2 ⇒ a → × b → = 3 Now we know vector product a → × b → is perpendicular to the plane of a → and b → . So the unit vector perpendicular to the plane of a → and b → is given as = ± a → × b → a → × b → Now substituting the corresponding value, we get = ± - i ^ - j ^ + k ^ 3 Or, the unit vector can be written as = ± - 1 3 i ^ - 1 3 j ^ + 1 3 k ^ .

Find a unit vector perpendicular to the plane of two vectors a → and b → , where a → = 4 i ^ - j ^ + 3 k ^ and b → = - 2 i ^ + j ^ - 2 k ^

Given, a → = 4 i ^ - j ^ + 3 k ^ and b → = - 2 i ^ + j ^ - 2 k ^ Now the cross product of these vectors will be a → × b → = i ^ j ^ k ^ 4 - 1 3 - 2 1 - 2 ⇒ a → × b → = 2 - 3 i ^ - - 8 + 6 j ^ + 4 - 2 k ^ ⇒ a → × b → = - i ^ + 2 j ^ + 2 k ^ And the corresponding magnitude will be a → × b → = - 1 2 + 2 2 + 2 2 ⇒ a → × b → = 9 = 3 Now we know vector product a → × b → will be perpendicular to the plane of a → and b → . So the unit vector perpendicular to the plane of a → and b → is given as = ± a → × b → a → × b → Now substituting the corresponding value, we get = ± - i ^ + 2 j ^ + 2 k ^ 3 Or, the unit vector can be written as = ± - 1 3 i ^ + 2 3 j ^ + 2 3 k ^ .

Find a unit vector perpendicular to the vectors 2 i ^ + 3 j ^ - k ^ and i ^ + 2 j ^ + 3 k ^

Let the two vectors given are a → = 2 i ^ + 3 j ^ - k ^ and b → = i ^ + 2 j ^ + 3 k ^ . Now the cross product of these vectors will be a → × b → = i ^ j ^ k ^ 2 3 - 1 1 2 3 ⇒ a → × b → = 9 + 2 i ^ - 6 + 1 j ^ + 4 - 3 k ^ ⇒ a → × b → = 11 i ^ - 7 j ^ + k ^ And the corresponding magnitude will be a → × b → = 11 2 + - 7 2 + 1 2 ⇒ a → × b → = 121 + 49 + 1 ⇒ a → × b → = 171 ⇒ a → × b → = 9 × 19 ⇒ a → × b → = 3 19 Now we know vector product a → × b → will be perpendicular to the plane of a → and b → . So the unit vector perpendicular to the plane of a → and b → is given as = ± a → × b → a → × b → Now substituting the corresponding value, we get = ± 11 i ^ - 7 j ^ + k ^ 3 19 Or, the unit vector can be written as = ± 11 3 19 i ^ - 7 3 19 j ^ + 1 3 19 k ^ .