Question:
The \(x\)-intercept of a plane \(\pi\) passing through the point \((1,1,1)\) is \(\dfrac{5}{2}\) and the perpendicular distance from the origin to the plane \(\pi\) is \(\dfrac{5}{7}\). If the \(y\)-intercept of the plane \(\pi\) is negative and the \(z\)-intercept is positive, then its \(y\)-intercept is
The \(x\)-intercept of a plane \(\pi\) passing through the point \((1,1,1)\) is \(\dfrac{5}{2}\) and the perpendicular distance from the origin to the plane \(\pi\) is \(\dfrac{5}{7}\). If the \(y\)-intercept of the plane \(\pi\) is negative and the \(z\)-intercept is positive, then its \(y\)-intercept is
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For a plane in intercept form \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\), use the point condition and distance formula together to determine unknown intercepts.
Updated On: Jun 15, 2026
- \(-\dfrac{5}{3}\)
- \(-\dfrac{5}{6}\)
- \(-\dfrac{3}{2}\)
- \(-\dfrac{5}{2}\)
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The Correct Option is A
Solution and Explanation
Step 1: Use intercept form of the plane.
The intercept form of a plane is \[ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 \] where \(a,b,c\) are the \(x\)-, \(y\)- and \(z\)- intercepts respectively.
Given, \[ a=\frac{5}{2} \] So, the equation becomes \[ \frac{x}{5/2}+\frac{y}{b}+\frac{z}{c}=1 \] \[ \frac{2x}{5}+\frac{y}{b}+\frac{z}{c}=1 \]
Step 2: Use the fact that the plane passes through \((1,1,1)\).
Substituting \((1,1,1)\) into the plane equation, \[ \frac{2}{5}+\frac{1}{b}+\frac{1}{c}=1 \] \[ \frac{1}{b}+\frac{1}{c}=1-\frac{2}{5} \] \[ \frac{1}{b}+\frac{1}{c}=\frac{3}{5} \]
Step 3: Use the perpendicular distance formula.
The plane equation is \[ \frac{2x}{5}+\frac{y}{b}+\frac{z}{c}-1=0 \] Distance of the origin from the plane is \[ \frac{| -1 |}{\sqrt{\left(\frac{2}{5}\right)^2+\left(\frac{1}{b}\right)^2+\left(\frac{1}{c}\right)^2}} =\frac{5}{7} \] Thus, \[ \frac{1}{\sqrt{\frac{4}{25}+\left(\frac{1}{b}\right)^2+\left(\frac{1}{c}\right)^2}} =\frac{5}{7} \] Squaring both sides, \[ \frac{4}{25}+\left(\frac{1}{b}\right)^2+\left(\frac{1}{c}\right)^2 =\frac{49}{25} \] \[ \left(\frac{1}{b}\right)^2+\left(\frac{1}{c}\right)^2 =\frac{45}{25} =\frac{9}{5} \]
Step 4: Solve for \(\dfrac{1}{b}\) and \(\dfrac{1}{c}\).
Let \[ u=\frac{1}{b},\qquad v=\frac{1}{c} \] Then, \[ u+v=\frac{3}{5} \] and \[ u^2+v^2=\frac{9}{5} \] Using \[ (u+v)^2=u^2+v^2+2uv \] \[ \left(\frac{3}{5}\right)^2=\frac{9}{5}+2uv \] \[ \frac{9}{25}=\frac{45}{25}+2uv \] \[ 2uv=-\frac{36}{25} \] \[ uv=-\frac{18}{25} \] Thus, \(u\) and \(v\) satisfy \[ t^2-\frac{3}{5}t-\frac{18}{25}=0 \] Multiplying by \(25\), \[ 25t^2-15t-18=0 \] Solving, \[ t=\frac{15\pm45}{50} \] Hence, \[ t=\frac{6}{5}\quad \text{or}\quad t=-\frac{3}{5} \] Therefore, \[ \frac{1}{b}=-\frac{3}{5},\qquad \frac{1}{c}=\frac{6}{5} \] Since the \(y\)-intercept is negative, \[ b=-\frac{5}{3} \]
Step 5: Final conclusion.
Hence, the required \(y\)-intercept is \[ \boxed{-\frac{5}{3}} \]
The intercept form of a plane is \[ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 \] where \(a,b,c\) are the \(x\)-, \(y\)- and \(z\)- intercepts respectively.
Given, \[ a=\frac{5}{2} \] So, the equation becomes \[ \frac{x}{5/2}+\frac{y}{b}+\frac{z}{c}=1 \] \[ \frac{2x}{5}+\frac{y}{b}+\frac{z}{c}=1 \]
Step 2: Use the fact that the plane passes through \((1,1,1)\).
Substituting \((1,1,1)\) into the plane equation, \[ \frac{2}{5}+\frac{1}{b}+\frac{1}{c}=1 \] \[ \frac{1}{b}+\frac{1}{c}=1-\frac{2}{5} \] \[ \frac{1}{b}+\frac{1}{c}=\frac{3}{5} \]
Step 3: Use the perpendicular distance formula.
The plane equation is \[ \frac{2x}{5}+\frac{y}{b}+\frac{z}{c}-1=0 \] Distance of the origin from the plane is \[ \frac{| -1 |}{\sqrt{\left(\frac{2}{5}\right)^2+\left(\frac{1}{b}\right)^2+\left(\frac{1}{c}\right)^2}} =\frac{5}{7} \] Thus, \[ \frac{1}{\sqrt{\frac{4}{25}+\left(\frac{1}{b}\right)^2+\left(\frac{1}{c}\right)^2}} =\frac{5}{7} \] Squaring both sides, \[ \frac{4}{25}+\left(\frac{1}{b}\right)^2+\left(\frac{1}{c}\right)^2 =\frac{49}{25} \] \[ \left(\frac{1}{b}\right)^2+\left(\frac{1}{c}\right)^2 =\frac{45}{25} =\frac{9}{5} \]
Step 4: Solve for \(\dfrac{1}{b}\) and \(\dfrac{1}{c}\).
Let \[ u=\frac{1}{b},\qquad v=\frac{1}{c} \] Then, \[ u+v=\frac{3}{5} \] and \[ u^2+v^2=\frac{9}{5} \] Using \[ (u+v)^2=u^2+v^2+2uv \] \[ \left(\frac{3}{5}\right)^2=\frac{9}{5}+2uv \] \[ \frac{9}{25}=\frac{45}{25}+2uv \] \[ 2uv=-\frac{36}{25} \] \[ uv=-\frac{18}{25} \] Thus, \(u\) and \(v\) satisfy \[ t^2-\frac{3}{5}t-\frac{18}{25}=0 \] Multiplying by \(25\), \[ 25t^2-15t-18=0 \] Solving, \[ t=\frac{15\pm45}{50} \] Hence, \[ t=\frac{6}{5}\quad \text{or}\quad t=-\frac{3}{5} \] Therefore, \[ \frac{1}{b}=-\frac{3}{5},\qquad \frac{1}{c}=\frac{6}{5} \] Since the \(y\)-intercept is negative, \[ b=-\frac{5}{3} \]
Step 5: Final conclusion.
Hence, the required \(y\)-intercept is \[ \boxed{-\frac{5}{3}} \]
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