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**careless25****Real Member**- Registered: 2008-07-24
- Posts: 560

I have a math mid-term tomorrow and I cant do these from the Review. Can you solve/explain them to me ASAP?

1. Determine the components of a vector of length 44 that lies on the line of intersection of the planes with equations 3x - 4y + 9z = 0 and 2y - 9z = 0.

2. The line through a point P(a,0,a) with direction vector (-1,2,-1) intersects the plane 3x + 5y + 2z = 0 at point Q. The line through P with direction vector (-3,2,-1) intersects the plane at point R. For what choice of a is the distance between Q and R equal to 3?

3. Consider the two lines:

L1: (x,y,z) = (2,0,0) + t(1,2,-1)

L2: (x,y,z) = (3,2,3) + s(a,b,1)

where s and t are real numbers. Find a relationship between a and b (independent of s and t) that ensures that L1 and L2 intersect.

For this one, I get the right answer but I m not sure if my method is correct.

I take the 2 points given and subtract them and find the direction vector (1,2,3).

I also take the 2 direction vectors given and subtract them to obtain (a-1,b-2,2).

I make x1=x2, y1=y2, z1=z2.

I get:

a-1 = 1

a = 2

b-2 = 2

b = 4

3 = 2??

and a = 1/2 b is the answer.

4. In the following system of equations, k is a real number.

-2x + 4y + z = k + 1

kx + z = 0

y + kz = 0

a. For what valuess of k does the system:

i) have no solutions?

ii) have exactly one solution?

iii) have an infinite amount of solutions?

Thank you for your help!

C25

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**Monox D. I-Fly****Member**- From: Indonesia
- Registered: 2015-12-02
- Posts: 2,000

Any idea how to answer number 4? I tried kx + z = y + kz and got this:

kx – y + z – kz = 0

kx – y + (1 – k)z = 0

To make the system have no solution or infinite amount of solutions, I think kx – y + (1 – k)z is a multiple of -2x + 4y + z thus I multiplied it by -4 and got -4kx + 4y + (-4 + 4k)z = 0. From the x coefficient, I got k = 1/2, but from the z coefficient I got 5/4. Different answers. Was what I do wrong? What is the proper way to solve it? Thanks.

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May his adventurous soul rest in peace at heaven.

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,299

hi Monox D. I-Fly

I thought 'who is this examiner to set the exam for a Saturday'? but then I noticed the post date. Let's hope he did well.

This is what I did for number 4:

Note: It's easy to eliminate x and y to get an equation in z.

2 .... gives x = -z/k and 3 ... gives y = -kz so substitute these into 1 to get an equation for z.

It has a quadratic denominator so the roots will lead to impossible to solve.

I used the function plotter https://www.mathsisfun.com/data/function-grapher.php? to see what values of z are possible. There are three asymptotes, two vertical at the roots mentioned above and one horizontal.

As this is a valid function, any value of k that leads to a (unique) value of z means a single set of values for z, x and y. That's most values of k. There are two values at approximately - 0.6 and +0.8 (if my algebra is correct) where z tends to infinity so no solutions. I cannot find any values of k where more than one solution exists. At one point in my algebra I multiplied through by k to remove the fraction so I separately tested k=0 and that leads to a single solution too.

Bob

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You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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