Question

6. Which of the following is a unit vector perpendicular to the plane determined by the vectors A = 2i + 4j and B = i + j - k? Ο Α. -2i+j-k ов. (i+2j) 1, ос. Taxi ) To(-2i+ j- -zi- To -j-k) (-2i

Answer 1

The unit vector perpendicular to the plane determined by vectors A = 2i + 4j and B = i + j - k is C = (-2i + j - k).

Step-by-step explanation:

To find a unit vector perpendicular to the plane formed by vectors A and B we first need to calculate the cross product of A and B. The cross product of two vectors denoted as A × B yields a vector that is perpendicular to the plane defined by A and B. In this case:

`\[ A * B = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & 0 \\ 1 & 1 & -1 \end{vmatrix} \]`

Expanding the determinant we get:

`\[ A * B = \hat{i}(4 \cdot (-1) - 0 \cdot 1) - \hat{j}(2 \cdot (-1) - 0 \cdot 1) + \hat{k}(2 \cdot 1 - 4 \cdot 1) \]`

Simplifying further, we obtain:

`\[ A * B = -4\hat{i} + 2\hat{j} - 2\hat{k} \]`

To convert this vector into a unit vector, we divide each component by its magnitude. The magnitude of the vector C = -4 2 -2 is given by:

`\[ |C| = √((-4)^2 + 2^2 + (-2)^2) = √(16 + 4 + 4) = √(24) \]`

So the unit vector is:

`\[ \hat{C} = (C)/(|C|) = ((-4 ,2 -2))/(√(24)) = (1)/(√(6))(-2\hat{i} + \hat{j} - \hat{k}) \]`

Therefore the unit vector perpendicular to the plane determined by vectors A and B is C = (-2i + j - k).

Answer 2

The unit vector perpendicular to the plane determined by vectors A and B is computed using the cross product. After obtaining the orthogonal vector, its magnitude is calculated and used to normalize it, resulting in the unit vector -2/√5 i + 1/√5 j - 1/√5 k.

Step-by-step explanation:

The question is asking to find a unit vector that is perpendicular to the plane determined by vectors A = 2i + 4j and B = i + j - k. To do this, we use the cross product of A and B since the cross product gives a vector that is orthogonal to both A and B.

Using the cross product, we calculate:

A × B = (2i + 4j) × (i + j - k)

≈ 2(i × i) + 2(i × j) + 2(i × -k) + 4(j × i) + 4(j × j) + 4(j × -k)

≈ 0 + 2k + 2j - 4k + 0 - 4i

≈ -4i + 2j - 2k

To convert this to a unit vector, we calculate its magnitude and divide each component by the magnitude.

Magnitude = √((-4)^2 + 2^2 + (-2)^2) = √(20) = 2√5

The unit vector is then:

(-4i + 2j - 2k) / (2√5) = -2/√5 i + 1/√5 j - 1/√5 k