Question Video: Equivalent Vectors on a Coordinate Plane | Nagwa Question Video: Equivalent Vectors on a Coordinate Plane | Nagwa

# Question Video: Equivalent Vectors on a Coordinate Plane Mathematics • First Year of Secondary School

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The points π΄, π΅, and πΆ have coordinates (β7, 1), (β2, 4), and (β4, β1) respectively. Given that π¨π© and ππ are equivalent vectors, find the coordinates of π·.

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### Video Transcript

The points π΄, π΅, and πΆ have coordinates negative seven, one; negative two, four; and negative four, negative one, respectively. Given that the vector from π΄ to π΅ and the vector from πΆ to π· are equivalent vectors, find the coordinates of π·.

In this question, weβre given the coordinates of three points, π΄, π΅, and πΆ, and we are told that the vector from π΄ to π΅ is equivalent to the vector from πΆ to π·. We need to use this information along with the given diagram of the coordinates of these points and the vector from π΄ to π΅ to determine the coordinates of π·.

To do this, letβs start by recalling what it means for two vectors to be equivalent. In actual fact, there are many ways of showing that two vectors are equivalent. We will recall the fact that for two vectors to be equivalent, they must have the same magnitude and direction. We could also check that they have the same dimension. However, weβre working in the coordinate plane. So this is not necessary, and this is usually checked when making sure the directions are equivalent.

In a sketch of a vector in a space, we can recall that the length of the line segment represents the magnitude of the vector and the direction of the vector is represented by the direction of the arrow. We want the vector from πΆ to π· to have the same magnitude and direction as the vector from π΄ to π΅. Since the magnitudes are equal, the line segment from πΆ to π· must be the same length as the line segment from π΄ to π΅. Similarly, since the directions are equal, the line between π΄ and π΅ must be parallel to the line from πΆ to π· and the vectors must point in the same direction.

We can use these properties to find the coordinates of point π·. We start by sketching a ray parallel to π΄π΅ starting at point πΆ. We know that π· must lie on this ray since the vector from πΆ to π· must have the same direction as the vector from π΄ to π΅. Next, we need the magnitudes of the vectors to be equivalent. So we choose point π· on the line so that the line segments π΄π΅ and πΆπ· have the same length. We could do this by translating line segment π΄π΅ onto πΆ such that the image of π΄ is coincident with πΆ or by sketching a circle centered at πΆ of radius π΄π΅. This gives us that the coordinates of π· are one, two.

This is not the only method we can use to determine the coordinates of π·. We can also represent the vectors in terms of their horizontal and vertical components. We recall that a vector in the plane can be represented by the horizontal and vertical displacements from its initial point to its terminal point, called its components. In the diagram, we can see that we move five units right and three units up when traveling from π΄ to π΅. This means that the horizontal displacement is positive five and the vertical displacement is positive three when traveling from π΄ to π΅. This allows us to write the vector from π΄ to π΅ in terms of its components as the vector five, three.

We can now recall that for two vectors to be equal, they must have the same dimension and all of the components must be equal. Therefore, the vector from πΆ to π· must be the vector five, three. This means that we can find the coordinates of π· by traveling five units right from πΆ and three units up. We see that this is the point one, two.

Therefore, we were able to show that for vector π¨π© to be equivalent to vector ππ, π· must have coordinates one, two.

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