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Lesson Video: Geometric Applications of Vectors Mathematics

In this video, we will learn how to use vector operations and vector properties to solve problems involving geometrical shapes.

15:22

Video Transcript

In this video, we will learn how to use vector operations and vector properties to solve problems involving geometric shapes. We will begin by recalling some key properties that we will use throughout this video.

When adding or subtracting two vectors, we add or subtract their individual components. We know that two vectors are parallel if they are scalar multiples of each other. The vectors 𝐀 and 𝐁 are parallel if vector 𝐀 is equal to π‘˜ multiplied by vector 𝐁, where π‘˜ is a constant not equal to zero. Two vectors are perpendicular if their scalar product is equal to zero. We can use these two properties to solve problems involving parallel and perpendicular lines in geometric shapes. We know that if two vectors are equal, they have the same magnitude and direction. If this is true for two sides of a geometric shape, the two sides will be parallel and equal in length.

We will now look at some examples, the first of which involves finding the coordinates of a rectangle.

𝐴𝐡𝐢𝐷 is a rectangle in which the coordinates of the points 𝐴, 𝐡, and 𝐢 are negative 18, negative two; negative 18, negative three; and negative eight, π‘˜, respectively. Use vectors to find the value of π‘˜ and the coordinates of point 𝐷.

One way of solving this problem would be to draw a rectangle on the coordinate plane; however, we are asked to use vectors. It therefore makes sense to consider some of the properties of a rectangle. A rectangle has two pairs of equal-length parallel sides. This means that the vector 𝐀𝐁 will be equal to the vector 𝐃𝐂. Likewise, the vector 𝐃𝐀 will be equal to the vector 𝐂𝐁. We also know that the angles in a rectangle are right angles. This means that vector 𝐀𝐁 is perpendicular to vector 𝐂𝐁. The same is true for the other sides that meet at right angles.

We recall that to calculate vector 𝐀𝐁, we subtract vector 𝐀 from vector 𝐁. In this question, vector 𝐀𝐁 is equal to negative 18, negative three minus negative 18, negative two. Negative 18 minus negative 18 is the same as negative 18 plus 18. This is equal to zero. Negative three minus negative two is equal to negative one. Therefore, vector 𝐀𝐁 equals zero, negative one. Vector 𝐂𝐁 is equal to vector 𝐁 minus vector 𝐂. This is equal to negative 18, negative three minus negative eight, π‘˜. This is equal to negative 10, negative three minus π‘˜.

We know that if two vectors are perpendicular, the scalar product equals zero. This means that the scalar product of 𝐀𝐁 and 𝐂𝐁 equals zero. Zero multiplied by negative 10 plus negative one multiplied by negative three minus π‘˜ is equal to zero. This simplifies to zero is equal to three plus π‘˜. Subtracting three from both sides of this equation gives us π‘˜ is equal to negative three. The value of π‘˜ is equal to negative three, which means that 𝐢 has coordinates negative eight, negative three.

If we let the coordinates of point 𝐷 be π‘₯, 𝑦, then vector 𝐃𝐂 is equal to negative eight, negative three minus π‘₯, 𝑦. This is equal to negative eight minus π‘₯, negative three minus 𝑦. As the vectors 𝐀𝐁 and 𝐃𝐂 have the same magnitude and direction, they must be equal. This means zero must be equal to negative eight minus π‘₯. Negative one must be equal to negative three minus 𝑦. Solving our first equation, we get π‘₯ is equal to negative eight. And solving the second equation, we get 𝑦 is equal to negative two. The coordinates of point 𝐷 are therefore equal to negative eight, negative two.

In our next question, we will consider a triangle.

In the triangle 𝐴𝐡𝐢, 𝐷 lies on the line segment 𝐡𝐢, where the ratio of 𝐡𝐷 to 𝐷𝐢 is two to three. Given that three multiplied by vector 𝐀𝐁 plus two multiplied by vector 𝐀𝐂 is equal to π‘˜ multiplied by vector 𝐀𝐃, find the value of π‘˜.

We will begin by considering the triangle 𝐴𝐡𝐢. We know that point 𝐷 lies on 𝐡𝐢 and the ratio of 𝐡𝐷 to 𝐷𝐢 is two to three. This means that the vector 𝐁𝐃 is two-fifths of the vector 𝐁𝐂. Let’s now consider the equation we’re given. Three multiplied by vector 𝐀𝐁 plus two multiplied by vector 𝐀𝐂 is equal to π‘˜ multiplied by vector 𝐀𝐃. We know that vector 𝐀𝐂 is equal to 𝐀𝐁 plus 𝐁𝐂. This means that the left-hand side of our equation can be rewritten as three 𝐀𝐁 plus two multiplied by 𝐀𝐁 plus 𝐁𝐂. We can distribute the parentheses to get two 𝐀𝐁 plus two 𝐁𝐂. Collecting like terms, we have five 𝐀𝐁 plus two 𝐁𝐂.

Let’s now consider the right-hand side of the equation. Vector 𝐀𝐃 is equal to 𝐀𝐁 plus 𝐁𝐃. Therefore, the right-hand side is equal to π‘˜ multiplied by 𝐀𝐁 plus 𝐁𝐃. We know that 𝐁𝐃 is equal to two-fifths of 𝐁𝐂. We can then distribute the parentheses. This gives us π‘˜ multiplied by vector 𝐀𝐁 plus two-fifths π‘˜ multiplied by vector 𝐁𝐂. We can now equate coefficients. Five must be equal to π‘˜. Two must be equal to two-fifths of π‘˜. Dividing both sides of this equation by two-fifths also gives us π‘˜ is equal to five. The value for π‘˜ such that three multiplied by vector 𝐀𝐁 plus two multiplied by vector 𝐀𝐂 is equal to π‘˜ multiplied by vector 𝐀𝐃 is five.

In our next question, we will look at the properties of a square.

𝐴𝐡𝐢𝐷 is a square in which the coordinates of the points 𝐴, 𝐡, and 𝐢 are one, negative eight; three, negative 10; and five, negative eight. Use vectors to determine the coordinates of point 𝐷 and the area of the square.

We know that a square has four equal-length sides and the angles are all right angles. We know that two vectors are equal if they have the same magnitude and direction. This means that, in this case, vector 𝐀𝐁 is equal to vector 𝐃𝐂. We know that vector 𝐀𝐁 is equal to vector 𝐁 minus vector 𝐀. In this question, we need to subtract the vector one, negative eight from the vector three, negative 10. Three minus one is equal to two, and negative 10 minus negative eight is equal to negative two.

We can repeat this for vector 𝐃𝐂 where the coordinates of point 𝐷 are π‘₯, 𝑦. We need to subtract π‘₯, 𝑦 from five, negative eight. This gives us five minus π‘₯ and negative eight minus 𝑦. As the vectors 𝐀𝐁 and 𝐃𝐂 are equal, we can now equate the components. Two is equal to five minus π‘₯, and negative two is equal to negative eight minus 𝑦. These equations can be rearranged so that π‘₯ is equal to five minus two and 𝑦 is equal to negative eight plus two. Five minus two equals three, and negative eight plus two is negative six. The coordinates of point 𝐷 are therefore equal to three, negative six.

In the second part of this question, we need to find the area of the square. In order to do this, we need to find the length of one of the sides. This is the same as the magnitude of the vector. The magnitude of any vector 𝐀𝐁 is equal to the square root of π‘₯ squared plus 𝑦 squared. We square root the sum of the squares of each of the components. In this question, the magnitude of 𝐀𝐁 is equal to the square root of two squared plus negative two squared. Two squared and negative two squared are both equal to four. The magnitude of vector 𝐀𝐁 is therefore equal to the square root of eight. This means that the length of each side of the square is the square root of eight.

We know that in order to calculate the area of a square, we square the side length, in this case, root eight squared. As squaring and square rooting are the inverse or opposite of each other, the area of the square is eight square units.

In our final question, we will look at a trapezoid or trapezium.

Trapezoid 𝐴𝐡𝐢𝐷 has vertices 𝐴: four, 14; 𝐡: four, negative four; 𝐢: negative 12, negative four; and 𝐷: negative 12, nine. Given that vector 𝐀𝐁 is parallel to vector 𝐃𝐂 and vector 𝐀𝐁 is perpendicular to vector 𝐂𝐁, find the area of that trapezoid.

Let’s consider what we are told about the trapezoid. We know that vector 𝐀𝐁 is parallel to vector 𝐃𝐂. We also know that vector 𝐀𝐁 is perpendicular to vector 𝐂𝐁. This means that they meet at right angles. It therefore follows that vector 𝐃𝐂 also meets vector 𝐂𝐁 at right angles. We know that the area of a trapezoid can be found using the formula π‘Ž plus 𝑏 over two multiplied by β„Ž, where π‘Ž and 𝑏 are the parallel sides and β„Ž is the perpendicular height. We need to find the length of the sides 𝐴𝐡 and 𝐷𝐢 and the perpendicular height 𝐢𝐡.

In order to calculate the length of the sides, we need to work out the magnitude of the vectors. We will begin by calculating the magnitude of 𝐀𝐁. This is equal to the square root of four minus four squared plus negative four minus 14 squared. Four minus four is equal to zero. And negative four minus 14 is negative 18. Squaring this gives us 324, and then square rooting the answer gives us 18. The magnitude of vector 𝐀𝐁 is 18. We can repeat this process to calculate the magnitude of vector 𝐃𝐂. This is equal to the square root of negative 12 minus negative 12 squared plus negative four minus nine squared. This gives us an answer of 13.

As the magnitude of 𝐀𝐁 is greater than the magnitude of 𝐃𝐂, we can see that our sketch has not been drawn to scale. It would therefore make more sense to relabel it as shown. Vector 𝐀𝐁 is still parallel to vector 𝐃𝐂, and vector 𝐀𝐁 is perpendicular to vector 𝐂𝐁. We can now add the lengths onto our diagram. We now need to calculate the magnitude of vector 𝐁𝐂. Using the same method, we see that this is equal to 16. We now have the lengths of the parallel sides as well as the length of the perpendicular height of the trapezoid. The area is therefore equal to 13 plus 18 divided by two multiplied by 16. 13 plus 18 is equal to 31. Multiplying 31 over two by 16 gives us 248. The area of the trapezoid is therefore equal to 248 square units.

We will now finish this video by summarizing the key points. We saw in this video that we can use vector operations and vector properties to solve problems involving geometric shapes. We use the fact that two vectors are parallel if they’re scalar multiples of each other. Vector 𝐀 is equal to π‘˜ multiplied by vector 𝐁. We also use the fact that two vectors are perpendicular if their scalar product is equal to zero. We also saw that if vector 𝐀 has components π‘₯, 𝑦, then the magnitude of vector 𝐀 is equal to the square root of π‘₯ squared plus 𝑦 squared. This enabled us to solve problems involving the area of squares, trapezoids, rectangles, and triangles.

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