Question Video: Finding the Value of an Expression Using Dot Product Mathematics

If 𝐀 = 3𝐒 + 2𝐣 βˆ’ 3𝐀, 𝐁 = 2𝐒 βˆ’ 2𝐣 βˆ’ 𝐀, and 𝐂 = 2𝐒 βˆ’ 𝐣 βˆ’ 3𝐀, find 2𝐂 βˆ’ [2𝐀 β‹… (2𝐁 + 4𝐂)]𝐁.

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Video Transcript

If vector 𝐀 is equal to three 𝐒 plus two 𝐣 minus three 𝐀, vector 𝐁 is equal to two 𝐒 minus two 𝐣 minus 𝐀, and vector 𝐂 is equal to two 𝐒 minus 𝐣 minus three 𝐀, find two 𝐂 minus the dot product of two 𝐀 and two 𝐁 plus four 𝐂 multiplied by vector 𝐁.

The expression we’re trying to calculate here is quite complicated. We will, therefore, begin by calculating the expression inside the parentheses, two 𝐁 plus four 𝐂. When multiplying any vector by a scalar or constant, we simply multiply each of the components by that scalar. This means that two 𝐁 is equal to two multiplied by two 𝐒 minus two 𝐣 minus 𝐀. This simplifies to four 𝐒 minus four 𝐣 minus two 𝐀.

We can calculate four 𝐂 in a similar way. This is equal to eight 𝐒 minus four 𝐣 minus 12𝐀. Our next step is to add these two vectors. When adding any two vectors, we simply add their corresponding components. Four 𝐒 plus eight 𝐒 is equal to 12𝐒, negative four 𝐣 plus negative four 𝐣 is equal to negative eight 𝐣, and negative two 𝐀 plus negative 12𝐀 is negative 14𝐀. Two 𝐁 plus four 𝐂 is equal to 12𝐒 minus eight 𝐣 minus 14𝐀.

Our next step is to find a dot product of two 𝐀 and two 𝐁 plus four 𝐂. Multiplying vector 𝐀 by two gives us six 𝐒 plus four 𝐣 minus six 𝐀. We calculate the dot or scalar product of any two vectors by multiplying the corresponding components and then finding the sum of these values. We begin by multiplying 12 and six. Next, we need to multiply negative eight and four. And finally, we multiply negative 14 and negative six. The dot product is equal to 12 multiplied by six plus negative eight multiplied by four plus negative 14 multiplied by negative six. This simplifies to 72 plus negative 32 plus 84, which in turn is equal to 124.

The entire expression inside the square brackets which is the dot product of two 𝐀 and two 𝐁 plus four 𝐂 is equal to the scalar or constant 124. Multiplying this by vector 𝐁, we have that 124𝐁 is equal to 248𝐒 minus 248𝐣 minus 124𝐀. We can now work out the final answer by subtracting this from two 𝐂. Two 𝐂 is equal to four 𝐒 minus two 𝐣 minus six 𝐀. Subtracting the 𝐒-components gives us negative 244𝐒. Subtracting the 𝐣-components gives us positive 246𝐣. Finally, subtracting the 𝐀-components gives us 118𝐀.

If vector 𝐀 is equal to three 𝐒 plus two 𝐣 minus three 𝐀, vector 𝐁 is equal to two 𝐒 minus two 𝐣 minus 𝐀, and vector 𝐂 is equal to two 𝐒 minus 𝐣 minus three 𝐀, then two 𝐂 minus the dot product of two 𝐀 and two 𝐁 plus four 𝐂 multiplied by vector 𝐁 is equal to negative 244𝐒 plus 246𝐣 plus 118𝐀.

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