Find the image of line |
\[\frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-\,5}\] in the plane \[2x-y+z+3=0.\] |
OR |
Find the distance of the point (3, 4, 5) from the plane \[x+y+z=2\] measured parallel to the line\[2x=y=z\]. |
Answer:
Given equation of the line is \[\frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-\,5}\] ?(i) and the given plane is \[2x-y+z+3=0\] The DR's of line (i) are \[(3,\,\,1,\,\,-5)\] and the DR's of normal to the plane (ii) are \[(2,\,\,-1,\,\,1).\] Now, by using \[{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0,\] we get \[(3)(2)+(1)(-1)+(-\,5)(1)=0\] \[\Rightarrow \] \[6-1-5=0\] \[\Rightarrow \] 0 = 0 Also, putting x =1, y = 3 and z = 4 in LHS of Eq. (ii), we get \[LHS=2(1)-(3)+(4)+3=2-3+4+3=6\ne 0\] \[\therefore \] The point P(1, 3, 4) of line (i) does not lie in the plane (ii). Hence, line (i) is parallel to the plane (ii). \[\therefore \] DR's of the line normal to plane (ii) is \[(2,\,\,-1,\,\,1).\] Let \[Q(\alpha ,\,\,\beta ,\,\,\gamma )\] be the image of the point P(1, 3, 4). Now, equation of line PM passing through (1, 3, 4) and normal to plane (ii) is \[\frac{x-1}{2}=\frac{y-3}{-\,1}=\frac{z-4}{1}=\lambda \,\,(say)\] \[\Rightarrow \] \[x=2\lambda +1,\] \[y=-\,\lambda +3,\] \[z=\lambda +4\] \[\therefore \] Any point M on the line \[=(2\lambda +1),\] \[(-\,\lambda +3),\]\[(\lambda +4)\] ?(iii) Since, the point M lies on the plane (ii). \[\therefore \] \[=2(2\lambda +1)-\,1(-\,\lambda +3)+\lambda +4+3=0\] \[\Rightarrow \] \[4\lambda +2+\lambda -3+\lambda +4+3=0\] \[\Rightarrow \] \[6\lambda +6=0\] \[\Rightarrow \] \[\lambda =-\,1\] On putting the value of \[\lambda \] in Eq. (iii), we get' Coordinates of \[m=(-\,2+1,\,\,1+3,\,\,-\,1+4)=(-\,1,\,\,4,\,\,3)\] Since, M is the mid-point of line PQ. \[\therefore \] \[(-\,1,\,\,4,\,\,3)=\left( \frac{\alpha +1}{2},\,\,\frac{\beta +3}{2},\,\,\frac{\gamma +4}{2} \right)\] Then, comparing the corresponding coordinate, we get \[\frac{\alpha +1}{2}=-\,1\] \[\Rightarrow \] \[\alpha +1=-\,2\] \[\Rightarrow \] \[\alpha =-\,3,\] \[\frac{\beta +3}{2}=4\] \[\Rightarrow \] \[\beta +3=8\] \[\Rightarrow \] \[\beta =8-3=5\] and \[\frac{\gamma +4}{2}=3\] \[\Rightarrow \] \[\gamma +4=6\] \[\Rightarrow \] \[\gamma =2\] \[\therefore \] Coordinates of \[Q=(-\,3,\,\,5,\,\,2)\] Hence, the equation of line through Q and parallel to the given line (i) is \[\frac{x+3}{3}=\frac{y-5}{1}=\frac{z-2}{-\,5},\] which is the required Image of line (i). OR Given, point = (3, 4, 5) Equation of plane is\[x+y+z=2\] ...(i) and equation of line is \[2x=y=z\] \[\Rightarrow \] \[\frac{x}{1}=\frac{y}{2}=\frac{z}{2}\] ?(ii) Let the equation of the line passing through the point (3, 4, 5) and parallel to the line (ii) be \[\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}=\lambda \,(say)\] ?(iii) Then, any point A on the line (iii) is \[A(\lambda +3,\,\,2\lambda +4,\,\,2\lambda +5)\] If the point \[A(\lambda +3,\,\,2\lambda +4,\,\,2\lambda +5)\] lies on the plane (i), then \[(\lambda +3)+(2\lambda +4)+(2\lambda +5)=2\] \[\Rightarrow \] \[5\lambda +12=2\,\,\,\,\,\Rightarrow \,\,\,\,\,5\lambda =-\,10\,\,\,\,\,\Rightarrow \,\,\,\,\,\lambda =-\,2\] On substituting the value of \[\lambda \] in Eq. (iv), we get the coordinates of the point \[A(-\,2+3,\,\,-\,4+4,-\,4+5),\] i.e. \[A(1,\,\,0,\,\,1)\] Now, distance between the points P(3, 4, 5)and A(1, 0, 1) \[=\sqrt{{{(3-1)}^{2}}+{{(4-0)}^{2}}+{{(5-1)}^{2}}}\] \[=\sqrt{4+16+16}\] \[=\sqrt{36}=6\] Hence, the required distance of the point P(3, 4, 5)from the plane \[x+y+z=2\] measured parallel to the line \[2x=y=z\] is 6 units.
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