Notes On Properties of Cross Product - CBSE Class 12 Maths
Property 1: $\stackrel{\to }{\text{p}}$  x   is also a vector. The cross product of $\stackrel{\to }{\text{p}}$  and   is given by: $\stackrel{\to }{\text{p}}$  x $\stackrel{\to }{\text{q}}$  = |$\stackrel{\to }{\text{p}}$ | | |  sin ϴ $\stackrel{^}{\text{n}}$ Hence, the cross product of two vectors p and q, and the unit vector, n, are collinear. Property 2: $\stackrel{\to }{\text{p}}$  x   = $\stackrel{\to }{\text{0}}$  ⇔ $\stackrel{\to }{\text{p}}$    ||   $\stackrel{\to }{\text{q}}$  . Let $\stackrel{\to }{\text{p}}$  and $\stackrel{\to }{\text{q}}$  be two non-zero vectors. If $\stackrel{\to }{\text{p}}$  ∥  , ϴ = 0o ⇒ $\stackrel{\to }{\text{p}}$  x $\stackrel{\to }{\text{q}}$  = |$\stackrel{\to }{\text{p}}$ | | |  sin 0 $\stackrel{^}{\text{n}}$  =   (sin 0o = 0) Observation 1: $\stackrel{\to }{\text{p}}$  x $\stackrel{\to }{\text{p}}$  =   Since ϴ = 0o Observation 2: $\stackrel{\to }{\text{p}}$  x ( - $\stackrel{\to }{\text{p}}$ ) =   Since ϴ = 180o Observation 3: x $\stackrel{^}{\text{i}}$ =   x $\stackrel{^}{\text{j}}$ = $\stackrel{\to }{\text{0}}$  x $\stackrel{^}{\text{k}}$ = $\stackrel{\to }{\text{0}}$  If $\stackrel{\to }{\text{p}}$  Á $\stackrel{\to }{\text{q}}$ , ϴ = 90o Property 3: Given $\stackrel{\to }{\text{p}}$  Á $\stackrel{\to }{\text{q}}$ , $\stackrel{\to }{\text{p}}$  x $\stackrel{\to }{\text{q}}$  =  |$\stackrel{\to }{\text{p}}$ | | |  sin 90 $\stackrel{^}{\text{n}}$ =  |$\stackrel{\to }{\text{p}}$ | | | $\stackrel{^}{\text{n}}$   ⇒ $\stackrel{\to }{\text{p}}$   x    =  |$\stackrel{\to }{\text{p}}$ | | | |$\stackrel{^}{\text{n}}$|  = |$\stackrel{\to }{\text{p}}$ | | |  (sin 90o = 1 and $\stackrel{^}{\text{n}}$ is a unit vector) Property 4: Angle ϴ between two vectors $\stackrel{\to }{\text{p}}$  and $\stackrel{\to }{\text{q}}$  is given by: sin ϴ = $\frac{\text{|}\stackrel{_}{\text{p}}\text{x}\stackrel{_}{\text{q}}\text{|}}{\text{|}\stackrel{_}{\text{p}}\text{| |}\stackrel{_}{\text{q}}\text{|}}$ $\stackrel{\to }{\text{p}}$  x $\stackrel{\to }{\text{q}}$  = |$\stackrel{\to }{\text{p}}$ | | | sin ϴ $\stackrel{^}{\text{n}}$ ...(1)   x $\stackrel{\to }{\text{p}}$  =  |$\stackrel{\to }{\text{p}}$ | | | sin ϴ $\stackrel{^}{\text{n}}$, ...(2) $\stackrel{^}{\text{n}}$ = - $\stackrel{^}{\text{n}}$   x $\stackrel{\to }{\text{p}}$  = - |$\stackrel{\to }{\text{p}}$ | | | sin ϴ $\stackrel{^}{\text{n}}$ ...(3) From (1) and (3): $\stackrel{\to }{\text{p}}$  x $\stackrel{\to }{\text{q}}$  = - $\stackrel{\to }{\text{q}}$   x $\stackrel{\to }{\text{p}}$  Property 5: The cross product of vectors is not commutative, since $\stackrel{\to }{\text{p}}$  x $\stackrel{\to }{\text{q}}$  = - $\stackrel{\to }{\text{q}}$   x $\stackrel{\to }{\text{p}}$ . For mutually perpendicular vectors $\stackrel{^}{\text{i}}$ and $\stackrel{^}{\text{j}}$: x $\stackrel{^}{\text{j}}$ = || || sin 90 $\stackrel{^}{\text{n}}$ ⇒ x $\stackrel{^}{\text{j}}$ = $\stackrel{^}{\text{n}}$ (since  || = || = sin 90 = 1) ⇒    x $\stackrel{^}{\text{j}}$ = (since $\stackrel{^}{\text{n}}$ ⊥ $\stackrel{^}{\text{i}}$   x $\stackrel{^}{\text{j}}$ ) Since cross product is not commutative: x $\stackrel{^}{\text{i}}$ = -   Similarly: x  $\stackrel{^}{\text{k}}$ =  $\stackrel{^}{\text{i}}$ and $\stackrel{^}{\text{k}}$ x = -  $\stackrel{^}{\text{i}}$ $\stackrel{^}{\text{k}}$ x $\stackrel{^}{\text{i}}$ = and $\stackrel{^}{\text{i}}$ x  $\stackrel{^}{\text{k}}$ = -   Property 6: If $\stackrel{\to }{\text{p}}$  and $\stackrel{\to }{\text{q}}$  represent the adjacent sides of a triangle, then the area of the triangle = ½ | $\stackrel{\to }{\text{p}}$  x  $\stackrel{\to }{\text{q}}$ |. In ∆OPQ, let QR Á OP. Area of ∆OPQ = ½ Base x Height ⇒ Area of ∆OPQ = ½ (OP). ( QR) ...(1) In right-angled ∆OQR: sin ϴ = $\frac{\text{QR}}{\text{OQ}}$ ⇒ QR = OQ sin ϴ ⇒ Area of ∆OPQ = ½ (OP) (OQ) sin ϴ OP = |$\stackrel{\to }{\text{p}}$ | and OQ = | | ⇒ Area of ∆OPQ = ½ |$\stackrel{\to }{\text{p}}$ | | | sin ϴ Since | $\stackrel{\to }{\text{p}}$  x  $\stackrel{\to }{\text{q}}$ | = |$\stackrel{\to }{\text{p}}$ | | | sin ϴ Area of ∆OPQ = ½ |$\stackrel{\to }{\text{p}}$  x $\stackrel{\to }{\text{q}}$ | The cross product of $\stackrel{\to }{\text{p}}$  and $\stackrel{\to }{\text{q}}$  is given by: $\stackrel{\to }{\text{p}}$  x $\stackrel{\to }{\text{q}}$  = |$\stackrel{\to }{\text{p}}$  | | | sin ϴ $\stackrel{^}{\text{n}}$ Property 7: If $\stackrel{\to }{\text{p}}$  and $\stackrel{\to }{\text{q}}$  represent adjacent sides of a parallelogram, the area of the parallelogram = |$\stackrel{\to }{\text{p}}$  x $\stackrel{\to }{\text{q}}$ |. In parallelogram OPRQ, let QS Á OP. Area of parallelogram OPRQ = Base x Height ⇒ Area of parallelogram OPRQ = OP .QS ...(1) In right-angled ∆OQS: sin ϴ = $\frac{\text{QS}}{\text{OQ}}$ ⇒ QS = OQ sin ϴ ⇒ Area of parallelogram OPRQ = OP . OQ sin ϴ OP = |$\stackrel{\to }{\text{p}}$ | and OQ = | | ⇒ Area of parallelogram OPRQ =  |$\stackrel{\to }{\text{p}}$ | | |  sin ϴ Since |$\stackrel{\to }{\text{p}}$  x $\stackrel{\to }{\text{q}}$ | = |$\stackrel{\to }{\text{p}}$ | | | sin ϴ Area of parallelogram OPRQ = |$\stackrel{\to }{\text{p}}$  x $\stackrel{\to }{\text{q}}$ | Property 7: Distributive property of cross product. For any three vectors $\stackrel{\to }{\text{p}}$  ,  and $\stackrel{\to }{\text{r}}$   : $\stackrel{\to }{\text{p}}$  x ( $\stackrel{\to }{\text{q}}$   + $\stackrel{\to }{\text{r}}$ ) =  $\stackrel{\to }{\text{p}}$  x    +  $\stackrel{\to }{\text{p}}$  x $\stackrel{\to }{\text{r}}$  Property 8: For any two vectors $\stackrel{\to }{\text{p}}$  and $\stackrel{\to }{\text{q}}$  and a scalar λ :   λ ($\stackrel{\to }{\text{p}}$  x $\stackrel{\to }{\text{q}}$ ) = (λ  $\stackrel{\to }{\text{p}}$ ) x   = $\stackrel{\to }{\text{p}}$  x (λ $\stackrel{\to }{\text{q}}$ )    =  a1$\stackrel{^}{\text{i}}$ + a2 $\stackrel{^}{\text{j}}$ + a3 $\stackrel{^}{\text{k}}$      $\stackrel{\to }{\text{b}}$  =  b1$\stackrel{^}{\text{i}}$ + b2 $\stackrel{^}{\text{j}}$ + b3 $\stackrel{^}{\text{k}}$      x $\stackrel{\to }{\text{b}}$  = ( a1$\stackrel{^}{\text{i}}$ + a2 $\stackrel{^}{\text{j}}$ + a3 $\stackrel{^}{\text{k}}$   ) x (b1$\stackrel{^}{\text{i}}$ + b2 $\stackrel{^}{\text{j}}$ + b3 $\stackrel{^}{\text{k}}$ ) ⇒   x $\stackrel{\to }{\text{b}}$  =  a1$\stackrel{^}{\text{i}}$  x (b1$\stackrel{^}{\text{i}}$ + b2 $\stackrel{^}{\text{j}}$ + b3 $\stackrel{^}{\text{k}}$ ) +   a2 $\stackrel{^}{\text{j}}$   x (b1$\stackrel{^}{\text{i}}$ + b2 $\stackrel{^}{\text{j}}$ + b3 $\stackrel{^}{\text{k}}$ ) + a3 $\stackrel{^}{\text{k}}$   x (b1$\stackrel{^}{\text{i}}$ + b2 $\stackrel{^}{\text{j}}$ + b3 $\stackrel{^}{\text{k}}$ )    = a1b1 ($\stackrel{^}{\text{i}}$  x $\stackrel{^}{\text{i}}$ ) + a1b2 ($\stackrel{^}{\text{i}}$  x $\stackrel{^}{\text{j}}$ ) + a1b3 ($\stackrel{^}{\text{i}}$  x $\stackrel{^}{\text{k}}$ ) + a2b1 ($\stackrel{^}{\text{j}}$  x $\stackrel{^}{\text{i}}$ ) + a2b2 ($\stackrel{^}{\text{j}}$  x $\stackrel{^}{\text{j}}$ ) + a2b3 ($\stackrel{^}{\text{j}}$  x $\stackrel{^}{\text{k}}$ ) + a3b1 ($\stackrel{^}{\text{k}}$  x $\stackrel{^}{\text{i}}$) + a3b2 ($\stackrel{^}{\text{k}}$  x $\stackrel{^}{\text{j}}$) + a3b3 ($\stackrel{^}{\text{k}}$  x $\stackrel{^}{\text{k}}$ ) We know:  $\stackrel{^}{\text{i}}$  x $\stackrel{^}{\text{i}}$  = $\stackrel{^}{\text{j}}$  x $\stackrel{^}{\text{j}}$    =  $\stackrel{^}{\text{k}}$  x $\stackrel{^}{\text{k}}$    = 0   $\stackrel{^}{\text{i}}$ x $\stackrel{^}{\text{j}}$ = $\stackrel{^}{\text{k}}$ and $\stackrel{^}{\text{j}}$ x $\stackrel{^}{\text{i}}$ = - $\stackrel{^}{\text{k}}$ $\stackrel{^}{\text{j}}$ x $\stackrel{^}{\text{k}}$ = $\stackrel{^}{\text{i}}$ and $\stackrel{^}{\text{k}}$ x $\stackrel{^}{\text{j}}$ = - $\stackrel{^}{\text{i}}$ $\stackrel{^}{\text{k}}$ x $\stackrel{^}{\text{i}}$ = $\stackrel{^}{\text{j}}$ and $\stackrel{^}{\text{i}}$ x $\stackrel{^}{\text{k}}$ = - $\stackrel{^}{\text{j}}$  ⇒   x $\stackrel{\to }{\text{b}}$  =  a1b2 ($\stackrel{^}{\text{i}}$  x $\stackrel{^}{\text{j}}$ ) + a1b3 ($\stackrel{^}{\text{i}}$  x $\stackrel{^}{\text{k}}$ ) + a2b1 ($\stackrel{^}{\text{j}}$  x $\stackrel{^}{\text{i}}$ )  + a2b3 ($\stackrel{^}{\text{j}}$  x $\stackrel{^}{\text{k}}$ ) + a3b1 ($\stackrel{^}{\text{k}}$  x $\stackrel{^}{\text{i}}$) + a3b2 ($\stackrel{^}{\text{k}}$  x $\stackrel{^}{\text{j}}$)      =   a1b2 $\stackrel{^}{\text{k}}$   - a1b3 $\stackrel{^}{\text{j}}$   - a2b1 $\stackrel{^}{\text{k}}$    + a2b3 $\stackrel{^}{\text{i}}$   + a3b1 $\stackrel{^}{\text{j}}$   - a3b2 $\stackrel{^}{\text{i}}$            ⇒ $\stackrel{\to }{\text{a}}$  x $\stackrel{\to }{\text{b}}$  = ( a2b3  - a3b2) $\stackrel{^}{\text{i}}$   - (a1b3 -  a3b1)$\stackrel{^}{\text{j}}$   + (a1b2$\stackrel{}{\text{}}$$\stackrel{}{\text{}}$  - a2b1) $\stackrel{^}{\text{k}}$   The cross product of two vectors can also be expressed as a determinant. ⇒  $\stackrel{\to }{\text{a}}$  x $\stackrel{\to }{\text{b}}$  = $\left|\begin{array}{ccc}\stackrel{^}{\text{i}}& \stackrel{^}{\text{j}}& \stackrel{^}{\text{k}}\\ {\text{a}}_{\text{1}}& {\text{a}}_{\text{2}}& {\text{a}}_{\text{3}}\\ {\text{b}}_{\text{1}}& {\text{b}}_{\text{2}}& {\text{b}}_{\text{3}}\end{array}\right|$

#### Summary

Property 1: $\stackrel{\to }{\text{p}}$  x   is also a vector. The cross product of $\stackrel{\to }{\text{p}}$  and   is given by: $\stackrel{\to }{\text{p}}$  x $\stackrel{\to }{\text{q}}$  = |$\stackrel{\to }{\text{p}}$ | | |  sin ϴ $\stackrel{^}{\text{n}}$ Hence, the cross product of two vectors p and q, and the unit vector, n, are collinear. Property 2: $\stackrel{\to }{\text{p}}$  x   = $\stackrel{\to }{\text{0}}$  ⇔ $\stackrel{\to }{\text{p}}$    ||   $\stackrel{\to }{\text{q}}$  . Let $\stackrel{\to }{\text{p}}$  and $\stackrel{\to }{\text{q}}$  be two non-zero vectors. If $\stackrel{\to }{\text{p}}$  ∥  , ϴ = 0o ⇒ $\stackrel{\to }{\text{p}}$  x $\stackrel{\to }{\text{q}}$  = |$\stackrel{\to }{\text{p}}$ | | |  sin 0 $\stackrel{^}{\text{n}}$  =   (sin 0o = 0) Observation 1: $\stackrel{\to }{\text{p}}$  x $\stackrel{\to }{\text{p}}$  =   Since ϴ = 0o Observation 2: $\stackrel{\to }{\text{p}}$  x ( - $\stackrel{\to }{\text{p}}$ ) =   Since ϴ = 180o Observation 3: x $\stackrel{^}{\text{i}}$ =   x $\stackrel{^}{\text{j}}$ = $\stackrel{\to }{\text{0}}$  x $\stackrel{^}{\text{k}}$ = $\stackrel{\to }{\text{0}}$  If $\stackrel{\to }{\text{p}}$  Á $\stackrel{\to }{\text{q}}$ , ϴ = 90o Property 3: Given $\stackrel{\to }{\text{p}}$  Á $\stackrel{\to }{\text{q}}$ , $\stackrel{\to }{\text{p}}$  x $\stackrel{\to }{\text{q}}$  =  |$\stackrel{\to }{\text{p}}$ | | |  sin 90 $\stackrel{^}{\text{n}}$ =  |$\stackrel{\to }{\text{p}}$ | | | $\stackrel{^}{\text{n}}$   ⇒ $\stackrel{\to }{\text{p}}$   x    =  |$\stackrel{\to }{\text{p}}$ | | | |$\stackrel{^}{\text{n}}$|  = |$\stackrel{\to }{\text{p}}$ | | |  (sin 90o = 1 and $\stackrel{^}{\text{n}}$ is a unit vector) Property 4: Angle ϴ between two vectors $\stackrel{\to }{\text{p}}$  and $\stackrel{\to }{\text{q}}$  is given by: sin ϴ = $\frac{\text{|}\stackrel{_}{\text{p}}\text{x}\stackrel{_}{\text{q}}\text{|}}{\text{|}\stackrel{_}{\text{p}}\text{| |}\stackrel{_}{\text{q}}\text{|}}$ $\stackrel{\to }{\text{p}}$  x $\stackrel{\to }{\text{q}}$  = |$\stackrel{\to }{\text{p}}$ | | | sin ϴ $\stackrel{^}{\text{n}}$ ...(1)   x $\stackrel{\to }{\text{p}}$  =  |$\stackrel{\to }{\text{p}}$ | | | sin ϴ $\stackrel{^}{\text{n}}$, ...(2) $\stackrel{^}{\text{n}}$ = - $\stackrel{^}{\text{n}}$   x $\stackrel{\to }{\text{p}}$  = - |$\stackrel{\to }{\text{p}}$ | | | sin ϴ $\stackrel{^}{\text{n}}$ ...(3) From (1) and (3): $\stackrel{\to }{\text{p}}$  x $\stackrel{\to }{\text{q}}$  = - $\stackrel{\to }{\text{q}}$   x $\stackrel{\to }{\text{p}}$  Property 5: The cross product of vectors is not commutative, since $\stackrel{\to }{\text{p}}$  x $\stackrel{\to }{\text{q}}$  = - $\stackrel{\to }{\text{q}}$   x $\stackrel{\to }{\text{p}}$ . For mutually perpendicular vectors $\stackrel{^}{\text{i}}$ and $\stackrel{^}{\text{j}}$: x $\stackrel{^}{\text{j}}$ = || || sin 90 $\stackrel{^}{\text{n}}$ ⇒ x $\stackrel{^}{\text{j}}$ = $\stackrel{^}{\text{n}}$ (since  || = || = sin 90 = 1) ⇒    x $\stackrel{^}{\text{j}}$ = (since $\stackrel{^}{\text{n}}$ ⊥ $\stackrel{^}{\text{i}}$   x $\stackrel{^}{\text{j}}$ ) Since cross product is not commutative: x $\stackrel{^}{\text{i}}$ = -   Similarly: x  $\stackrel{^}{\text{k}}$ =  $\stackrel{^}{\text{i}}$ and $\stackrel{^}{\text{k}}$ x = -  $\stackrel{^}{\text{i}}$ $\stackrel{^}{\text{k}}$ x $\stackrel{^}{\text{i}}$ = and $\stackrel{^}{\text{i}}$ x  $\stackrel{^}{\text{k}}$ = -   Property 6: If $\stackrel{\to }{\text{p}}$  and $\stackrel{\to }{\text{q}}$  represent the adjacent sides of a triangle, then the area of the triangle = ½ | $\stackrel{\to }{\text{p}}$  x  $\stackrel{\to }{\text{q}}$ |. In ∆OPQ, let QR Á OP. Area of ∆OPQ = ½ Base x Height ⇒ Area of ∆OPQ = ½ (OP). ( QR) ...(1) In right-angled ∆OQR: sin ϴ = $\frac{\text{QR}}{\text{OQ}}$ ⇒ QR = OQ sin ϴ ⇒ Area of ∆OPQ = ½ (OP) (OQ) sin ϴ OP = |$\stackrel{\to }{\text{p}}$ | and OQ = | | ⇒ Area of ∆OPQ = ½ |$\stackrel{\to }{\text{p}}$ | | | sin ϴ Since | $\stackrel{\to }{\text{p}}$  x  $\stackrel{\to }{\text{q}}$ | = |$\stackrel{\to }{\text{p}}$ | | | sin ϴ Area of ∆OPQ = ½ |$\stackrel{\to }{\text{p}}$  x $\stackrel{\to }{\text{q}}$ | The cross product of $\stackrel{\to }{\text{p}}$  and $\stackrel{\to }{\text{q}}$  is given by: $\stackrel{\to }{\text{p}}$  x $\stackrel{\to }{\text{q}}$  = |$\stackrel{\to }{\text{p}}$  | | | sin ϴ $\stackrel{^}{\text{n}}$ Property 7: If $\stackrel{\to }{\text{p}}$  and $\stackrel{\to }{\text{q}}$  represent adjacent sides of a parallelogram, the area of the parallelogram = |$\stackrel{\to }{\text{p}}$  x $\stackrel{\to }{\text{q}}$ |. In parallelogram OPRQ, let QS Á OP. Area of parallelogram OPRQ = Base x Height ⇒ Area of parallelogram OPRQ = OP .QS ...(1) In right-angled ∆OQS: sin ϴ = $\frac{\text{QS}}{\text{OQ}}$ ⇒ QS = OQ sin ϴ ⇒ Area of parallelogram OPRQ = OP . OQ sin ϴ OP = |$\stackrel{\to }{\text{p}}$ | and OQ = | | ⇒ Area of parallelogram OPRQ =  |$\stackrel{\to }{\text{p}}$ | | |  sin ϴ Since |$\stackrel{\to }{\text{p}}$  x $\stackrel{\to }{\text{q}}$ | = |$\stackrel{\to }{\text{p}}$ | | | sin ϴ Area of parallelogram OPRQ = |$\stackrel{\to }{\text{p}}$  x $\stackrel{\to }{\text{q}}$ | Property 7: Distributive property of cross product. For any three vectors $\stackrel{\to }{\text{p}}$  ,  and $\stackrel{\to }{\text{r}}$   : $\stackrel{\to }{\text{p}}$  x ( $\stackrel{\to }{\text{q}}$   + $\stackrel{\to }{\text{r}}$ ) =  $\stackrel{\to }{\text{p}}$  x    +  $\stackrel{\to }{\text{p}}$  x $\stackrel{\to }{\text{r}}$  Property 8: For any two vectors $\stackrel{\to }{\text{p}}$  and $\stackrel{\to }{\text{q}}$  and a scalar λ :   λ ($\stackrel{\to }{\text{p}}$  x $\stackrel{\to }{\text{q}}$ ) = (λ  $\stackrel{\to }{\text{p}}$ ) x   = $\stackrel{\to }{\text{p}}$  x (λ $\stackrel{\to }{\text{q}}$ )    =  a1$\stackrel{^}{\text{i}}$ + a2 $\stackrel{^}{\text{j}}$ + a3 $\stackrel{^}{\text{k}}$      $\stackrel{\to }{\text{b}}$  =  b1$\stackrel{^}{\text{i}}$ + b2 $\stackrel{^}{\text{j}}$ + b3 $\stackrel{^}{\text{k}}$      x $\stackrel{\to }{\text{b}}$  = ( a1$\stackrel{^}{\text{i}}$ + a2 $\stackrel{^}{\text{j}}$ + a3 $\stackrel{^}{\text{k}}$   ) x (b1$\stackrel{^}{\text{i}}$ + b2 $\stackrel{^}{\text{j}}$ + b3 $\stackrel{^}{\text{k}}$ ) ⇒   x $\stackrel{\to }{\text{b}}$  =  a1$\stackrel{^}{\text{i}}$  x (b1$\stackrel{^}{\text{i}}$ + b2 $\stackrel{^}{\text{j}}$ + b3 $\stackrel{^}{\text{k}}$ ) +   a2 $\stackrel{^}{\text{j}}$   x (b1$\stackrel{^}{\text{i}}$ + b2 $\stackrel{^}{\text{j}}$ + b3 $\stackrel{^}{\text{k}}$ ) + a3 $\stackrel{^}{\text{k}}$   x (b1$\stackrel{^}{\text{i}}$ + b2 $\stackrel{^}{\text{j}}$ + b3 $\stackrel{^}{\text{k}}$ )    = a1b1 ($\stackrel{^}{\text{i}}$  x $\stackrel{^}{\text{i}}$ ) + a1b2 ($\stackrel{^}{\text{i}}$  x $\stackrel{^}{\text{j}}$ ) + a1b3 ($\stackrel{^}{\text{i}}$  x $\stackrel{^}{\text{k}}$ ) + a2b1 ($\stackrel{^}{\text{j}}$  x $\stackrel{^}{\text{i}}$ ) + a2b2 ($\stackrel{^}{\text{j}}$  x $\stackrel{^}{\text{j}}$ ) + a2b3 ($\stackrel{^}{\text{j}}$  x $\stackrel{^}{\text{k}}$ ) + a3b1 ($\stackrel{^}{\text{k}}$  x $\stackrel{^}{\text{i}}$) + a3b2 ($\stackrel{^}{\text{k}}$  x $\stackrel{^}{\text{j}}$) + a3b3 ($\stackrel{^}{\text{k}}$  x $\stackrel{^}{\text{k}}$ ) We know:  $\stackrel{^}{\text{i}}$  x $\stackrel{^}{\text{i}}$  = $\stackrel{^}{\text{j}}$  x $\stackrel{^}{\text{j}}$    =  $\stackrel{^}{\text{k}}$  x $\stackrel{^}{\text{k}}$    = 0   $\stackrel{^}{\text{i}}$ x $\stackrel{^}{\text{j}}$ = $\stackrel{^}{\text{k}}$ and $\stackrel{^}{\text{j}}$ x $\stackrel{^}{\text{i}}$ = - $\stackrel{^}{\text{k}}$ $\stackrel{^}{\text{j}}$ x $\stackrel{^}{\text{k}}$ = $\stackrel{^}{\text{i}}$ and $\stackrel{^}{\text{k}}$ x $\stackrel{^}{\text{j}}$ = - $\stackrel{^}{\text{i}}$ $\stackrel{^}{\text{k}}$ x $\stackrel{^}{\text{i}}$ = $\stackrel{^}{\text{j}}$ and $\stackrel{^}{\text{i}}$ x $\stackrel{^}{\text{k}}$ = - $\stackrel{^}{\text{j}}$  ⇒   x $\stackrel{\to }{\text{b}}$  =  a1b2 ($\stackrel{^}{\text{i}}$  x $\stackrel{^}{\text{j}}$ ) + a1b3 ($\stackrel{^}{\text{i}}$  x $\stackrel{^}{\text{k}}$ ) + a2b1 ($\stackrel{^}{\text{j}}$  x $\stackrel{^}{\text{i}}$ )  + a2b3 ($\stackrel{^}{\text{j}}$  x $\stackrel{^}{\text{k}}$ ) + a3b1 ($\stackrel{^}{\text{k}}$  x $\stackrel{^}{\text{i}}$) + a3b2 ($\stackrel{^}{\text{k}}$  x $\stackrel{^}{\text{j}}$)      =   a1b2 $\stackrel{^}{\text{k}}$   - a1b3 $\stackrel{^}{\text{j}}$   - a2b1 $\stackrel{^}{\text{k}}$    + a2b3 $\stackrel{^}{\text{i}}$   + a3b1 $\stackrel{^}{\text{j}}$   - a3b2 $\stackrel{^}{\text{i}}$            ⇒ $\stackrel{\to }{\text{a}}$  x $\stackrel{\to }{\text{b}}$  = ( a2b3  - a3b2) $\stackrel{^}{\text{i}}$   - (a1b3 -  a3b1)$\stackrel{^}{\text{j}}$   + (a1b2$\stackrel{}{\text{}}$$\stackrel{}{\text{}}$  - a2b1) $\stackrel{^}{\text{k}}$   The cross product of two vectors can also be expressed as a determinant. ⇒  $\stackrel{\to }{\text{a}}$  x $\stackrel{\to }{\text{b}}$  = $\left|\begin{array}{ccc}\stackrel{^}{\text{i}}& \stackrel{^}{\text{j}}& \stackrel{^}{\text{k}}\\ {\text{a}}_{\text{1}}& {\text{a}}_{\text{2}}& {\text{a}}_{\text{3}}\\ {\text{b}}_{\text{1}}& {\text{b}}_{\text{2}}& {\text{b}}_{\text{3}}\end{array}\right|$

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