### Video Transcript

Given that point π΄ has coordinates zero, zero, five, express vector π¨π in terms of the unit vectors π’, π£, and π€.

Letβs begin by considering the three-dimensional coordinate plane as shown. We are told that point π΄ has coordinates zero, zero, five. This means that its π₯-coordinate is zero, its π¦-coordinate is zero, and its π§-coordinate is five. The point, therefore, lies on the π§-axis. The unit vectors π’ hat, π£ hat, and π€ hat are the vectors of magnitude one in the positive π₯-, π¦-, and π§-directions.

In this question, we need to express the vector from point π΄ to point π, where π is the origin with coordinate zero, zero, zero. To travel from point π΄ to point π, we travel zero units in the π₯-direction, zero units in the π¦-direction, and negative five units in the π§-direction. The vector π¨π is therefore equal to zero π’ plus zero π£ plus negative five π€. This simplifies to negative five π€.

An alternative method here would be to find the vector ππ¨ first. Any vector starting at the origin will have components equal to the coordinates of the endpoint. As point π΄ has coordinates zero, zero, five, the vector ππ¨ will be equal to zero π’ plus zero π£ plus five π€. This simplifies to five π€. We know that reversing the direction of a vector does not alter its magnitude. Vector π¨π is equal to the negative of vector ππ¨. This confirms, as we have already shown, that vector π¨π is equal to negative five π€.