Question Video: Determining the Length of the Perpendicular Drawn from a Given Point to a Straight Line | Nagwa Question Video: Determining the Length of the Perpendicular Drawn from a Given Point to a Straight Line | Nagwa

Question Video: Determining the Length of the Perpendicular Drawn from a Given Point to a Straight Line Mathematics • Third Year of Secondary School

Join Nagwa Classes

Determine, to the nearest hundredth, the length of the perpendicular drawn from the point (−5, −7, −10) to the straight line (𝑥 + 8)/2 = (𝑦 − 9)/8 = (𝑧 + 7)/−8.

04:20

Video Transcript

Determine, to the nearest hundredth, the length of the perpendicular drawn from the point negative five, negative seven, negative 10 to the straight line 𝑥 plus eight over two equals 𝑦 minus nine over eight equals 𝑧 plus seven over negative eight.

Okay, so here we have this known point in space and we also know there’s a straight line passing through space. And we want to know the length of the perpendicular from the line to the point. We’ll call this length 𝑑. We can recall that 𝑑 is given by this expression, where 𝐬 and 𝐏𝐋 are vectors. The condition for vector 𝐬 is that it is parallel to our given line. Regarding vector 𝐏𝐋, this is a vector that points from our given point in space, and we’ll call this point 𝑃, to a point that lies somewhere anywhere along our line. And we’ll call this point 𝐿. This then is what that vector looks like. We can see that, to solve for the distance 𝑑, we’ll need to know the components of a vector parallel to our line, the coordinates of a point on the line, and the coordinates of a point in space.

We’re given the coordinates of point 𝑃. And to solve for 𝐬 and 𝐿, we’ll use the equation of our straight line given here. What we want to do is to convert the form of this equation into what’s called vector form. It’s given to us in what is called symmetric form. And we can write it this way because all three of these fractions are equal to a scale factor we can call 𝑡. This means that we can write out separate equations for the 𝑥-, 𝑦-, and 𝑧-coordinates of our line. For example, the fact that 𝑥 plus eight over two equals 𝑡 means that 𝑥 must be equal to two times 𝑡 minus eight. Likewise, since 𝑦 minus nine over eight equals 𝑡, we can say that 𝑦 equals eight 𝑡 plus nine. And lastly, 𝑧 plus seven over negative eight equaling 𝑡 implies that 𝑧 equals negative eight 𝑡 minus seven.

Now, our line is expressed in what is called parametric form. And with just a few small changes, we can write this in vector form. If we collect the 𝑥, 𝑦, and 𝑧 equations together, we can say the 𝑥-, 𝑦-, and 𝑧-components of our line are equal to a vector negative eight, nine, negative seven plus our scale factor 𝑡 times the vector two, eight, negative eight. If we collect these 𝑥-, 𝑦-, and 𝑧-components into a single vector we’ll call 𝐫, then 𝐫 equals the vector negative eight, nine, negative seven plus the scale factor 𝑡 times the vector two, eight, negative eight.

This first vector in our equation is a vector from the origin of a coordinate frame to a point on the line. That tells us then that the point with coordinates negative eight, nine, negative seven lies along our line. And then over here, this vector we multiply by 𝑡 is a vector that runs along the length of our line. In other words, it’s parallel to it.

All this means we know the components of a vector parallel to our line and the coordinates of a point on it. The vector 𝐬 has components two, eight, negative eight, and the point 𝐿 has coordinates negative eight, nine, negative seven. Knowing this, we can now define the components of vector 𝐏𝐋 in our scenario. It’s equal to the coordinates of point 𝐿 minus those of point 𝑃. We get a vector with components negative three, 16, three.

Now that we know the components of 𝐬 and 𝐏𝐋, we’re ready to calculate the cross product of these vectors. That’s equal to the determinant of this three-by-three matrix. Here, in our first row are the 𝐢, 𝐣, and 𝐤 hat unit vectors and in the second and third rows the corresponding components of vectors 𝐏𝐋 and 𝐬, respectively. This is equal to 𝐢 hat times negative 152 minus 𝐣 hat times 18 plus 𝐤 hat times negative 56 or, in other words, a vector with components negative 152, negative 18, negative 56.

We’re now ready to calculate the magnitude of this cross product and then divide it by the magnitude of 𝐬. The magnitude of 𝐏𝐋 cross 𝐬 equals the square root of negative 152 squared plus negative 18 squared plus negative 56 squared. We then divide that by the magnitude of 𝐬, itself equal to the square root of two squared plus eight squared plus negative eight squared. When we evaluate this entire fraction here, the result, rounded to the nearest hundredth, is 14.19. This then is the length of the perpendicular from the given point to the given line.

Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions