# Question Video: Finding a Missing Value Using Vectors Mathematics

In the triangle 𝐴𝐵𝐶, 𝐷 ∈ line segment 𝐵𝐶, where 𝐵𝐷 : 𝐷𝐶 = 2 : 3. Given that 3𝐀𝐁 + 2𝐀𝐂 = 𝑘𝐀𝐃, find the value of 𝑘.

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### Video Transcript

In the triangle 𝐴𝐵𝐶, 𝐷 lies on the line segment 𝐵𝐶, where the ratio of 𝐵𝐷 to 𝐷𝐶 is two to three. Given that three multiplied by vector 𝐀𝐁 plus two multiplied by vector 𝐀𝐂 is equal to 𝑘 multiplied by vector 𝐀𝐃, find the value of 𝑘.

We will begin by considering the triangle 𝐴𝐵𝐶. We know that point 𝐷 lies on 𝐵𝐶 and the ratio of 𝐵𝐷 to 𝐷𝐶 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 𝑘 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 𝑘 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 𝑘 multiplied by vector 𝐀𝐁 plus two-fifths 𝑘 multiplied by vector 𝐁𝐂. We can now equate coefficients. Five must be equal to 𝑘. Two must be equal to two-fifths of 𝑘. Dividing both sides of this equation by two-fifths also gives us 𝑘 is equal to five. The value for 𝑘 such that three multiplied by vector 𝐀𝐁 plus two multiplied by vector 𝐀𝐂 is equal to 𝑘 multiplied by vector 𝐀𝐃 is five.