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Question: A point P lies on a line through \[Q\left( 1,-2,3 \right)\] and is parallel to the line \[\dfrac{x}{...

A point P lies on a line through Q(1,−2,3)Q\left( 1,-2,3 \right) and is parallel to the line x1=y4=z5\dfrac{x}{1}=\dfrac{y}{4}=\dfrac{z}{5}. If P lies on the plane 2x+3y−4z+22=02x+3y-4z+22=0, then the segment PQ is equals to
(a) 42\sqrt{42} units
(b) 32\sqrt{32} units
(c) 4 units
(d) 5 units

Explanation

Solution

We Solve this problem by considering the point P as (a,b,c)\left( a,b,c \right). Then, we use the first condition that is P lies on a line through Q(1,−2,3)Q\left( 1,-2,3 \right) and is parallel to the line x1=y4=z5\dfrac{x}{1}=\dfrac{y}{4}=\dfrac{z}{5} and find the line equation of PQ as when a line formed by A(x1,y1,z1)A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right) and B(x2,y2,z2)B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right) is parallel to xa=yb=zc\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}, then the equation of AB is given asx1−x2a=y1−y2b=z1−z2c\dfrac{{{x}_{1}}-{{x}_{2}}}{a}=\dfrac{{{y}_{1}}-{{y}_{2}}}{b}=\dfrac{{{z}_{1}}-{{z}_{2}}}{c}. Then, we use the second condition that P lies on plane 2x+3y−4z+22=02x+3y-4z+22=0 to get the point P. then we use the distance formula between two points as
PQ=(x2−x1)2+(y2−y1)2+(z2−z1)2PQ=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}

Complete step by step answer:
We are given the point Q as Q(1,−2,3)Q\left( 1,-2,3 \right).
Let us assume the required point P as P(a,b,c)P\left( a,b,c \right).
We are given that the point P lies on line through Q that is parallel to x1=y4=z5\dfrac{x}{1}=\dfrac{y}{4}=\dfrac{z}{5}.
We know that when a line formed by A(x1,y1,z1)A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right) and B(x2,y2,z2)B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right) is parallel to xa=yb=zc\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}, then the equation of AB is given asx1−x2a=y1−y2b=z1−z2c\dfrac{{{x}_{1}}-{{x}_{2}}}{a}=\dfrac{{{y}_{1}}-{{y}_{2}}}{b}=\dfrac{{{z}_{1}}-{{z}_{2}}}{c}.
By using the above equation let us find the equation of PQ as follows
⇒a−11=b+24=c−35\Rightarrow \dfrac{a-1}{1}=\dfrac{b+2}{4}=\dfrac{c-3}{5}
Let us assume the above line equation equals to ′k′'k'that is
⇒a−11=b+24=c−35=k\Rightarrow \dfrac{a-1}{1}=\dfrac{b+2}{4}=\dfrac{c-3}{5}=k
Now, let us take the each equation equals to ′k′'k' that is

& \Rightarrow \dfrac{a-1}{1}=k \\\ & \Rightarrow a=k+1 \\\ \end{aligned}$$ Similarly let us take the second equation as $$\begin{aligned} & \Rightarrow \dfrac{b+2}{4}=k \\\ & \Rightarrow b=4k-2 \\\ \end{aligned}$$ Now, let us take the third equation as $$\begin{aligned} & \Rightarrow \dfrac{c-3}{5}=k \\\ & \Rightarrow c=5k+3 \\\ \end{aligned}$$ We are given that the point lies on the plane $$2x+3y-4z+22=0$$ so that the point P satisfies the above plane equation that is $$\Rightarrow 2a+3b-4c+22=0$$ Now, by substituting the values of $$a,b,c$$ in above equation we get $$\begin{aligned} & \Rightarrow 2\left( k+1 \right)+3\left( 4k-2 \right)-4\left( 5k+3 \right)+22=0 \\\ & \Rightarrow 2k+2+12k-6-20k-12+22=0 \\\ & \Rightarrow -6k+6=0 \\\ & \Rightarrow k=1 \\\ \end{aligned}$$ Now, by substituting the value of $$'k'$$ in $$a,b,c$$ we get $$\begin{aligned} & \Rightarrow a=k+1=2 \\\ & \Rightarrow b=4k-2=2 \\\ & \Rightarrow c=5k+3=8 \\\ \end{aligned}$$ So, we can say that the point P is $$P\left( 2,2,8 \right)$$. We know that the distance between two points is given by the formula $$PQ=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}$$ By using the above formula to PQ we get $$\begin{aligned} & \Rightarrow PQ=\sqrt{{{\left( 2-1 \right)}^{2}}+{{\left( 2+2 \right)}^{2}}+{{\left( 8-3 \right)}^{2}}} \\\ & \Rightarrow PQ=\sqrt{1+16+25}=\sqrt{42} \\\ \end{aligned}$$ Therefore, the segment length PQ is $$\sqrt{42}$$. **So, the correct answer is “Option a”.** **Note:** Students may make mistakes in finding the equation of line PQ. The equation of line formed by $$A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$$ and $$B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$$ is parallel to $$\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}$$, then the equation of AB is given as$$\dfrac{{{x}_{1}}-{{x}_{2}}}{a}=\dfrac{{{y}_{1}}-{{y}_{2}}}{b}=\dfrac{{{z}_{1}}-{{z}_{2}}}{c}$$. But, students may do mistake in taking the equation formula as $$\dfrac{{{x}_{1}}-{{x}_{2}}}{a-{{x}_{2}}}=\dfrac{{{y}_{1}}-{{y}_{2}}}{b-{{y}_{2}}}=\dfrac{{{z}_{1}}-{{z}_{2}}}{c-{{z}_{2}}}$$ This is because they make mistakes in taking the equation $$\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}$$ as a line equation. If we are given that the line PQ is parallel to point then the above equation is correct. So, reading the question and applying the required formula is important in this question.