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Question: Find equation of line through the point \( (1,\; - 1,\;0) \) to intersect the lines \( {L_1} = \dfra...

Find equation of line through the point (1,  −1,  0)(1,\; - 1,\;0) to intersect the lines L1=x−22=y−13=z−34  and  L2=x−44=y5=z+15{L_1} = \dfrac{{x - 2}}{2} = \dfrac{{y - 1}}{3} = \dfrac{{z - 3}}{4}\;{\text{and}}\;{L_2} = \dfrac{{x - 4}}{4} = \dfrac{y}{5} = \dfrac{{z + 1}}{5}

Explanation

Solution

Hint : To find equation of line through given point and which also intersect two lines, you should know that equation of a line passing through two points (x1,  y1,  z1)  and  (x2,  y2,  z2)\left( {{x_1},\;{y_1},\;{z_1}} \right)\;{\text{and}}\;\left( {{x_2},\;{y_2},\;{z_2}} \right) can be written as x−x1x2−x1=y−y1y2−y1=z−z1z2−z1\dfrac{{x - {x_1}}}{{{x_2} - {x_1}}} = \dfrac{{y - {y_1}}}{{{y_2} - {y_1}}} = \dfrac{{z - {z_1}}}{{{z_2} - {z_1}}} , you have given one of the two points, and have to calculate coordinates of another point. Find another point by finding the intersection point of the given lines which are intersecting with the given line.

Complete step-by-step answer :
In order to find the equation of the line through the point (1,  −1,  0)(1,\; - 1,\;0) we have to find another point lying on the line. And for this, we will find the intersection point of the two given lines, that are L1=x−22=y−13=z−34  and  L2=x−44=y5=z+15{L_1} = \dfrac{{x - 2}}{2} = \dfrac{{y - 1}}{3} = \dfrac{{z - 3}}{4}\;{\text{and}}\;{L_2} = \dfrac{{x - 4}}{4} = \dfrac{y}{5} = \dfrac{{z + 1}}{5} because the line is also intersecting simultaneously with these lines.
We can find the point of intersection of these two lines as
L1=x−22=y−13=z−34=p  and  L2=x−44=y5=z+15=q ⇒x−22=p,  y−13=p,  z−34=p  and  ⇒x−44=q,  y5=q,  z+15=q ⇒x=2p+2,  y=3p+1,  z=4p+3  and  ⇒x=4q+4,  y=5q,  z=5q−1  {L_1} = \dfrac{{x - 2}}{2} = \dfrac{{y - 1}}{3} = \dfrac{{z - 3}}{4} = p\;{\text{and}}\;{L_2} = \dfrac{{x - 4}}{4} = \dfrac{y}{5} = \dfrac{{z + 1}}{5} = q \\\ \Rightarrow \dfrac{{x - 2}}{2} = p,\;\dfrac{{y - 1}}{3} = p,\;\dfrac{{z - 3}}{4} = p\;{\text{and}}\; \Rightarrow \dfrac{{x - 4}}{4} = q,\;\dfrac{y}{5} = q,\;\dfrac{{z + 1}}{5} = q \\\ \Rightarrow x = 2p + 2,\;y = 3p + 1,\;z = 4p + 3\;{\text{and}}\; \Rightarrow x = 4q + 4,\;y = 5q,\;z = 5q - 1 \\\
Now taking these two sets of coordinates values as the coordinate second point and then we can write,
x−12p+2−1=y−(−1)3p+1−(−1)=z−04p+3−0  and  x−14q+4−1=y−(−1)5q−(−1)=z−05q−1−0 x−12p+1=y+13p+2=z4p+3  and  x−14q+3=y+15q+1=z5q−1   \dfrac{{x - 1}}{{2p + 2 - 1}} = \dfrac{{y - ( - 1)}}{{3p + 1 - ( - 1)}} = \dfrac{{z - 0}}{{4p + 3 - 0}}\;{\text{and}}\;\dfrac{{x - 1}}{{4q + 4 - 1}} = \dfrac{{y - ( - 1)}}{{5q - ( - 1)}} = \dfrac{{z - 0}}{{5q - 1 - 0}} \\\ \dfrac{{x - 1}}{{2p + 1}} = \dfrac{{y + 1}}{{3p + 2}} = \dfrac{z}{{4p + 3}}\;{\text{and}}\;\dfrac{{x - 1}}{{4q + 3}} = \dfrac{{y + 1}}{{5q + 1}} = \dfrac{z}{{5q - 1}} \;
Since both the equations are representing equation of the same line, so we can write

⇒2p+14q+3=3p+25q+1=4p+35q−1=k(say) ⇒2p+14q+3=k,  3p+25q+1=k,  4p+35q−1=k ⇒2p+1=4qk+3k−−(i),  3p+2=5qk+k−−(ii),  4p+3=5qk−k(iii)   \Rightarrow \dfrac{{2p + 1}}{{4q + 3}} = \dfrac{{3p + 2}}{{5q + 1}} = \dfrac{{4p + 3}}{{5q - 1}} = k({\text{say}}) \\\ \Rightarrow \dfrac{{2p + 1}}{{4q + 3}} = k,\;\dfrac{{3p + 2}}{{5q + 1}} = k,\;\dfrac{{4p + 3}}{{5q - 1}} = k \\\ \Rightarrow 2p + 1 = 4qk + 3k - - (i),\;3p + 2 = 5qk + k - - (ii),\;4p + 3 = 5qk - k(iii) \;

From equation (iii)−(ii),(iii) - (ii), we get
⇒p+1=−2k−−−(iv)\Rightarrow p + 1 = - 2k - - - (iv)
From equation 4×(ii)−5×(i),4 \times (ii) - 5 \times (i), we get
⇒2p+3=−11k−−−(v)\Rightarrow 2p + 3 = - 11k - - - (v)
Again from (v)−2×(i),(v) - 2 \times (i), we get
k=−17k = - \dfrac{1}{7}
Putting k=−17k = - \dfrac{1}{7} in equation (iv),
⇒p=27−1=−57\Rightarrow p = \dfrac{2}{7} - 1 = - \dfrac{5}{7}
Putting p=−57p = - \dfrac{5}{7} in above equation, to get required equation,
⇒x−12(−57)+1=y+13(−57)+2=z4(−57)+3 ⇒7x−7−10+7=7y+7−15+14=7z−20+21 ⇒7x−7−3=7y+7−1=7z1 ⇒x−13=y+11=−z1   \Rightarrow \dfrac{{x - 1}}{{2\left( { - \dfrac{5}{7}} \right) + 1}} = \dfrac{{y + 1}}{{3\left( { - \dfrac{5}{7}} \right) + 2}} = \dfrac{z}{{4\left( { - \dfrac{5}{7}} \right) + 3}} \\\ \Rightarrow \dfrac{{7x - 7}}{{ - 10 + 7}} = \dfrac{{7y + 7}}{{ - 15 + 14}} = \dfrac{{7z}}{{ - 20 + 21}} \\\ \Rightarrow \dfrac{{7x - 7}}{{ - 3}} = \dfrac{{7y + 7}}{{ - 1}} = \dfrac{{7z}}{1} \\\ \Rightarrow \dfrac{{x - 1}}{3} = \dfrac{{y + 1}}{1} = - \dfrac{z}{1} \;
Therefore x−13=y+11=−z1\dfrac{{x - 1}}{3} = \dfrac{{y + 1}}{1} = - \dfrac{z}{1} is the equation of the required line.
So, the correct answer is “ x−13=y+11=−z1\dfrac{{x - 1}}{3} = \dfrac{{y + 1}}{1} = - \dfrac{z}{1} ”.

Note : This type of questions, have a lot of calculations, so do cross check your every calculation because one incorrect calculation will lead you to the wrong solution. Also when solving for the values of considered constants, you will have three variables and three equations, so you can either solve it by substitution method or with the help of matrices.