Question

Given two vectors A⃗ =−2.00i^+3.00j^+4.00k^ and B⃗ =3.00i^+1.00j^−3.00k^. Obtain a unit vector perpendicular to these two vectors.

Answer 1

To obtain a unit vector perpendicular to the given vectors A⃗ and B⃗ , we can calculate the cross product. The cross product A⃗ × B⃗ is a vector perpendicular to both A⃗ and B⃗ . By dividing the cross product vector by its magnitude, we can obtain a unit vector.

Step-by-step explanation:

To obtain a unit vector perpendicular to vectors A⃗ = -2.00i^ + 3.00j^ + 4.00k^ and B⃗ = 3.00i^ + 1.00j^ - 3.00k^, we can calculate the cross product of the two vectors. The cross product A⃗ × B⃗ is a vector that is perpendicular to both A⃗ and B⃗ . To get a unit vector, we divide the cross product vector by its magnitude.

Using the formula for the cross product, A⃗ × B⃗ = (AyBz - AzBy)i^ - (AxBz - AzBx)j^ + (AxBy - AyBx)k^, we can substitute the values and calculate: (-9.00)i^ - (3.00)j^ + (5.00)k^.

The magnitude of the cross product vector is |A⃗ × B⃗ | = √((-9.00)2 + (-3.00)2 + (5.00)2) = √115.

Finally, we can find the unit vector by dividing the cross product vector by its magnitude: (-9.00/√115)i^ - (3.00/√115)j^ + (5.00/√115)k^.

Answer 2

Step-by-step explanation:

Given

`\vec{A}=-2\hat{i}+3\hat{j}+4\hat{k}`

`\vec{B}=3\hat{i}+1\hat{j}-3\hat{k}`

Cross of two Product gives a vector which is perpendicular to the both vectors

`\vec{A}* \vec{B}=\begin{vmatrix}i &j &k \\ -2 &3 &4 \\ 3 &1 &-3 \end{vmatrix}`

`\vec{A}* \vec{B}=\hat{i}(-9-4)-\hat{j}(6-12)+\hat{k}(-2-9)`

`\vec{A}* \vec{B}=\vec{r}=-13\hat{i}+6\hat{j}-11\hat{k}`

Now unit vector in this direction is given by

`\hat{n}=\frac{\vec{r}}{|\vec{r}|}=\frac{-13\hat{i}+6\hat{j}-11\hat{k}}{√(13^2+6^2+11^2)}`

`\hat{n}=\frac{-13\hat{i}+6\hat{j}-11\hat{k}}{√(326)}`