Question

A vector perpendicular to both the vectors 2i-j+5k and x axis

Answer 1

Given:

The vectors 2i-j+5k and x axis

To find:

A vector perpendicular to both the vectors 2i-j+5k and x axis

Solution:

If two vectors are perpendicular, then their cross-product is defined to be A × B = (a2_b3 - a3_b2, a3_b1 - a1_b3, a1_b2 - a2*b1).

The cross product of two non-parallel vectors is a vector that is perpendicular to both of them.

(a1, a2, a3) = (2, -1, 5)

(b1, b2, b3) = (1, 0, 0)     (represents x-axis = 1i + 0j + 0k)

$\begin{pmatrix}2&-1&5\end{pmatrix}\times \begin{pmatrix}1&0&0\end{pmatrix}$

$=\begin{pmatrix}-1\cdot \:0-5\cdot \:0&5\cdot \:1-2\cdot \:0&2\cdot \:0-\left(-1\cdot \:1\right)\end{pmatrix}$

$=\begin{pmatrix}0&5&1\end{pmatrix}$

= 0i + 5j + k

∴ The vector perpendicular to both the vectors 2i-j+5k and x axis is 5j + k.

Answer 2

Step-by-step explanation:

If two vectors are perpendicular, then their cross-product is defined to be A × B = (a2_b3 - a3_b2, a3_b1 - a1_b3, a1_b2 - a2*b1).

The cross product of two non-parallel vectors is a vector that is perpendicular to both of them.

(a1, a2, a3) = (2, -1, 5)

(b1, b2, b3) = (1, 0, 0) (represents x-axis = 1i + 0j + 0k)

\begin{pmatrix}2&-1&5\end{pmatrix}\times \begin{pmatrix}1&0&0\end{pmatrix}(

2

−1

5

)×(

1

0

0

)

=\begin{pmatrix}-1\cdot \:0-5\cdot \:0&5\cdot \:1-2\cdot \:0&2\cdot \:0-\left(-1\cdot \:1\right)\end{pmatrix}=(

−1⋅0−5⋅0

5⋅1−2⋅0

2⋅0−(−1⋅1)

)

=\begin{pmatrix}0&5&1\end{pmatrix}=(

0

5

1

)

= 0i + 5j + k

∴ The vector perpendicular to both the vectors 2i-j+5k and x axis is 5j + k