Video: EG19M2-ALGANDGEO-Q11

Find the distance from the point (1, 5, βˆ’4) to the plane 2π‘₯ + 𝑦 βˆ’ 2𝑧 = 0, giving your answer in length units.

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Video Transcript

Find the distance from the point one, five, negative four to the plane two π‘₯ plus 𝑦 minus two 𝑧 equals zero, giving your answer in length units.

So we have a point one, five, negative four, which we can think of as the point 𝐴, and a plane with equation two π‘₯ plus 𝑦 minus two 𝑧 equals zero. And we’re asked to find the distance from the point 𝐴 to the plane.

Now when we’re asked to find the distance between a point and a plane, we are always talking about the perpendicular distance. We can do this by first finding any point on the plane, which we can call 𝐡. And we consider the vector from this point on the plane to our point. That’s the vector 𝐡𝐴, which I’ve marked in orange.

Now this vector won’t necessarily be perpendicular to the plane, which means its length won’t give us the distance between the point and the plane that we’re looking for. Instead, we can find a normal vector to the plane, which we’ll call the vector 𝑛. And we can place this anywhere that we like on the plane.

We can then find the projection of our orange vector, which is 𝐡𝐴, onto this normal vector 𝑛. And as this will be a multiple of the normal vector, it will be perpendicular to the plane. The length of this projection vector will then be the perpendicular distance from our point 𝐴 to the plane.

So what we’re looking for then is the projection of the vector 𝐡𝐴 onto our normal vector 𝑛. And first, we need to find the vectors 𝐡𝐴 and 𝑛. Let’s find a possibility for our point 𝐡 first of all, which, remember, was any point on the plane. For simplicity, let’s find the point where this plane intersects with the 𝑧-axis. The coordinates of this point will be zero, zero, 𝑧. Substituting these coordinates into the equation of the plane gives two multiplied by zero plus zero minus two 𝑧 is equal to zero. This just simplifies to negative two 𝑧 equals zero. And dividing both sides of this equation by negative two, we see that 𝑧 itself is also equal to zero. So in fact, our plane intersects with the 𝑧-axis at the origin. So our point 𝐡 has the coordinates zero, zero, zero.

Next, we need to find the vector 𝐡𝐴, which will just be the position vector of 𝐴 minus the position vector of 𝐡. The position vectors of these two points are just their coordinates. So we have one, five, negative four minus zero, zero, zero. We then subtract the component parts of these two vectors to give the vector with components one, five, negative four. In fact, as our point 𝐡 is just the origin, we find that the vector 𝐡𝐴 is actually just the position vector of point 𝐴.

Next, we need to consider the normal vector 𝑛. Now the π‘₯-, 𝑦-, and 𝑧-components of this normal vector are just the coefficients of π‘₯, 𝑦, and 𝑧 in the equation of the plane. The coefficient of π‘₯ is two, the coefficient of 𝑦 is one, and the coefficient of 𝑧 is negative two. So our normal vector has components two, one, negative two.

Now that we know the vectors 𝐡𝐴 and 𝑛, we can work out the projection of the vector 𝐡𝐴 onto 𝑛, which is given by the dot product of 𝐡𝐴 and 𝑛 over the dot product of 𝑛 and 𝑛 multiplied by the vector 𝑛. Remember that we’re looking for the length of this projection vector. So we need to find its magnitude. The fraction 𝐡𝐴 dot 𝑛 over 𝑛 dot 𝑛 is just a scalar. So we can take its absolute value outside of this magnitude.

In the denominator, the dot product of 𝑛 and 𝑛 is just the magnitude of 𝑛 squared. So in fact, one factor of the magnitude of 𝑛 will cancel with the magnitude of 𝑛 outside this fraction. The magnitude of our projection vector then simplifies to the absolute value of the dot product of the vectors 𝐡𝐴 and 𝑛 divided by the magnitude of the vector 𝑛.

Now let’s clear some space so we’ve got room to work this out. Our vector 𝐡𝐴 had components one, five, negative four. And our normal vector 𝑛 had components two, one, negative two. In the numerator then, we have one multiplied by two, which is two, plus five multiplied by one, which is five, plus negative four multiplied by negative two, which is eight. And in the denominator, the magnitude of the vector two, one, negative two is the square root of the sum of the squares of its individual components. That’s the square root of two squared plus one squared plus negative two squared. Two plus five plus eight is 15. And then in the denominator, two squared is four. Adding one squared gives five. And then adding negative two squared, so that’s adding another lot of four, gives nine.

So we have the absolute value of 15 over the square root of nine. The square root of nine is three, and 15 divided by three is five. The absolute value of five is just five. So the length of the projection vector of 𝐡𝐴 onto 𝑛 is five length units.

We found then that the distance by which we mean the perpendicular distance from the point one, five, negative four to the plane two π‘₯ plus 𝑦 minus two 𝑧 equals zero is five length units. We did this by first finding the coordinates of any point on the plane, which in this case were the coordinates of the 𝑧-intercept. We then found the vector from this point to our point 𝐴 and then the projection of this vector onto the normal vector to the plane. The magnitude of this projection vector gave us the perpendicular distance from the point to the plane.

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