Question

Find the equation of the line parallel to the vector 2i - j + 2k which passes through the point A whose position vector is 3i +j - k. If P is a point on this line such that AP = 15, find the position vector of P. ​

Answer 1

Answer:

Step-by-step explanation:

The equation of line which passes through the point A(3,1,-1) and parallel to the vector 2i-j+2k is

r= 3i+j-k+λ(2i-j+2k)

Therefore any point on the line be (2t-3,-t-1,2t+1)

Let the coordinate of P be ( 2t - 3,-t-1,2t+1)

The distance between A and P is  

According To Problem,

=15

⇒9-12t-181=0

⇒t=6,-3.4

Therefore the coordinate of P are (9,-7,13) and (-9.8,2.4,-5.8)

Answer 2

Answer:

The equation of line is:

$\frac{x-3}{2}= \frac{y-1}{-1} = \frac{z+1}{2}$

The position vector P is either 13i -4j + 9k or -7i +6j - 11k.

Step-by-step explanation:

Given,

The vector (V) 2i - j + 2k which is parallel to the unknown vector (U) and passes through the position vector A(3i + j - k)

And AP = 15

To find,

The equation of the line and the position vector of P.

Calculation,

For a line parallel to ai + bj + ck and passing through x'i + y'j + z'k, the line equation is:

$\frac{x - x'}{a} = \frac{y - y'}{b} = \frac{z - z'}{c}$

Here a = 2, b = -1, c = 2, and x' = 3, y' = 1, and z' = -1

Hence, the equation of the line is:

$\frac{x-3}{2}= \frac{y-1}{-1} = \frac{z-(-1)}{2} \\\implies \frac{x-3}{2}= \frac{y-1}{-1} = \frac{z+1}{2}$

Now we find the point P

We get that point P as:

P(2r + 3, -r + 1, 2r - 1) and A(3, 1, -1)

As AP = 15

⇒ (2r)² + (-r)² + (2r)² = 15²

⇒ 4r² + r² + 4r² = 225

⇒ 9r² = 225

⇒ r = ±5

If r = 5 then,

The position vector of P is 13i -4j + 9k.

If r = -5 then,

The position vector of P is -7i +6j - 11k

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