WEBVTT
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the question they asked me to find the symmetry equations
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for the line of intersections of these planes, the
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number one is equal to two, X minus by
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minus five. And play number two is equal to
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four. Express three by minus five. Yeah,
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so the equations of the planes can be re adding
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some regular formats. Two X minus one, man
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is equal to five. And second plane is four
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X plus three by men. Is that equal to
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five? So the normal vectors of each of these
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planes and one vector is equal to 2-1-1
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and and two vector is equal to 4, 3
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and-1. And so we have to find the
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perpendicular of the line of intersection of these planes.
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So and vector is equal to anyone victor across into
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victor. And according to the cross production we find
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the value of n vector is equal to four,
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icap minus two, Jacob plus 10-K cup. Therefore
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Normal vector to the line of intersection of this plane
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is equal to 4-2 and 10. Yeah.
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Okay. Yeah. So after finding the normal victor
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to the point of intersection on this plane, we
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have to find out point of intersection of these planes
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coordinates. Mhm. So in order to find the
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coordinates we have to put an arbitrary value To any
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variable, let zero equal to zero. Therefore the
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first equation becomes two weeks minus y equal to five
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. And the second equation becomes four X last three
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way Is equal to five. So if we calculate
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the values of X and y. From this equations
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we multiply the first equation by three we get uh
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the coefficients for x will be six X-3,
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Y is equal to 15. Therefore, if we
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add district questions, why 10 will get cancelled and
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the question will become 10, X is equal to
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20. Therefore from here X is equal to two
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. So from the above equation, if we put
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the value of X, we can get the value
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of Why? So four into 2 plus three,
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Y equal to five. From the second equation,
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we can find the value of Y. So three
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Y is equal to five minus eight. Therefore Why
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is equal to-1. We have found out coordinates
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of the point of intersection of these two planes.
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So the coordinates are mhm Okay, Mhm. 2
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-1 and zero. Now we have to find this
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symmetric equation at the line of intersection of these planes
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that can be evaluated as follows. So, mm
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hmm. So for finding this imagery equation of the
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line of intersection of these two planes, we have
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found out the normal vector, the point of intersection
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of these two planes that is equal to 4-2
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in 10. And the point of coordinate at the
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point of intersection of these two planes is equal to
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2-1 and zero. And therefore the symmetric equation
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can originals x minus the point according it Or x
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coordinate-2 divided by the normal vector for this coordinate
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, that is four is equal to why minus the
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point of coordinate for the y axis, that is
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minus one divided by The normal vector corresponding to this
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coordinate. That is-2 is equal to zero access
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minus the point of coordinates, or this access divided
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by the corresponding normal victor Value, that is 10
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. So after simplifying the situation it is equal to
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x minus two by four is equal to Y plus
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one by minus two is equal to zed by 10
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. And hence this is the symmetric equation at the
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line of intersection of the given planes.