Consider the line which passes through the point P(-1, -5, 4), and which is parallel to the line x=1+5t y=2+5t z=3+5t. How do you find the point of intersection of this new line with each of the coordinate planes?

Answer 1

The reqd. pts. of intersection are,

#{(-5,-9,0)}, {(0,-4,5)} and {(4,0,9)}#.

We first find the eqn. of the new line, say #L.#

We note that the Direction Vector of the given line is,

# (5,5,5)=5(1,1,1)#
Since the line #L# is parallel to this given one, we may take, as its
direction vector, #vec l=(1,1,1).# Also, #P(-1,-5,4) in L.#
Hence, the Cartesian Parametric Eqns. of #L# are,
# {x-(-1)}/1={y-(-5)}/1=(z-4)/1=k, say, where, k in RR; or,#
# L : x=k-1, y=k-5, z=k+4, k in RR... ... ... ...(star)#
Now, to find the pt. of int. of #L# with #3# co-ordinate planes, namely,
#(1)# the XY co-ord. pln., #(2)# the YZ pln. and, #(3)# the XZ pln.,

let us recall that, their resp. eqns. are,

#(1) :z=0; (2) : x=0, & (3) : y=0.#
Therefore, to find, #L nn XY"-plane, we solve "(star) & (1).#
Sub.ing, #z=0# in #(star)#, we get,
#k=-4, so, x=k-1=-4-1=-5, &, y=-9.#
# rArr L nn XY-#plane=#{(-5,-9,0)}.#
Similarly, #L nn YZ"-plane="{(0,-4,5)}...[because, k=1], and, #
# L nn XZ"-plane="{(4,0,9)}...[because, k=5].#

Thus, the reqd. pts. of intersection are,

#{(-5,-9,0)}, {(0,-4,5)} and {(4,0,9)}#.

Enjoy Maths.!

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Answer 2

To find the point of intersection of the line passing through point P(-1, -5, 4) and parallel to the line (x=1+5t), (y=2+5t), (z=3+5t) with each of the coordinate planes, follow these steps:

  1. Intersection with the xy-plane (z = 0): Substitute (z = 0) into the equation of the line to find the corresponding (x) and (y) coordinates. Then solve for (t). [ 4 + 5t = 0 \Rightarrow t = -\frac{4}{5} ] Substitute (t = -\frac{4}{5}) into the equations of the line to find (x) and (y).

  2. Intersection with the xz-plane (y = 0): Substitute (y = 0) into the equation of the line to find the corresponding (x) and (z) coordinates. Then solve for (t). [ -5 + 5t = 0 \Rightarrow t = 1 ] Substitute (t = 1) into the equations of the line to find (x) and (z).

  3. Intersection with the yz-plane (x = 0): Substitute (x = 0) into the equation of the line to find the corresponding (y) and (z) coordinates. Then solve for (t). [ -1 + 5t = 0 \Rightarrow t = \frac{1}{5} ] Substitute (t = \frac{1}{5}) into the equations of the line to find (y) and (z).

This process yields the points of intersection of the given line with each of the coordinate planes.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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