Question Video: Solving for Vector Components Using Parallel and Perpendicular Vector Relationships | Nagwa Question Video: Solving for Vector Components Using Parallel and Perpendicular Vector Relationships | Nagwa

# Question Video: Solving for Vector Components Using Parallel and Perpendicular Vector Relationships Mathematics • First Year of Secondary School

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### Video Transcript

If π¨ equals 12, eight; π© equals three, π; π equals π, nine; and π¨ is parallel to π©, π© is perpendicular to π, then π plus π equals what.

So, here we have these three vectors π¨, π©, and π, and weβre told that π¨ and π© are parallel while π© and π are perpendicular. As a side note, this means that π¨ is also perpendicular to π. But we wonβt need to use that fact to solve for π and π, these components of vectors π© and π, respectively. We can start out by solving for the value of π in vector π©. Now, the fact that this vector is parallel to vector π¨ means that we can write, for some nonzero constant πΆ, that πΆ times π© equals π¨. Another way of saying this is that the π₯-component of π΄ over the π₯-component of π΅ equals that same ratio for the corresponding π¦-components of these vectors.

From the given information, we see that π΄ sub π₯ equals 12, π΅ sub π₯ equals three, π΄ sub π¦ equals eight, while π΅ sub π¦ is π. So, 12 over three equals eight over π. And if we multiply both sides of this equation by three and by π, we find that 12 times π equals three times eight or 24. Dividing both sides of this equation by 12, we find that π equals two. So, we now know part of our answer of what π plus π will be. We can now use this information and the fact that π© is perpendicular to vector π to solve for π.

In general, if two vectors, weβll call them π one and π two, are perpendicular to one another, that means their dot product is zero. So, since π© is perpendicular to π, we can write that π© dot π equals zero, and therefore three, π dot π, nine equals zero. Before we calculate this dot product, letβs substitute in our known value for π. We know π is two, so three, two dot π, nine equals zero. And now, carrying out this dot product by first multiplying corresponding components of these vectors, we find that three times π plus two times nine is zero so that subtracting 18 from both sides, we have that three times π equals negative 18. Dividing both sides by three, we find that π equals negative six. We have then our values of π and π, and adding them together, we get a result of negative four.

So, if π¨ equals 12, eight; π© equals three, π; π equals π, nine; and π¨ is parallel to π©, π© is perpendicular to π, then π plus π equals negative four.

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