Question:
If
\[
-2,\ \frac{4}{3},\ -\frac{4}{5}
\]
are the intercepts made by a plane on \(X,Y,Z\)-axes respectively, then the direction cosines of a normal to this plane are:
If
\[
-2,\ \frac{4}{3},\ -\frac{4}{5}
\]
are the intercepts made by a plane on \(X,Y,Z\)-axes respectively, then the direction cosines of a normal to this plane are:
Show Hint
If a plane is given in intercept form
\[
\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1,
\]
first convert it into standard form \(Ax+By+Cz+D=0\). Then \((A,B,C)\) gives a normal vector to the plane.
Updated On: Jun 24, 2026
- \(\left(-\dfrac{1}{3},\dfrac{2}{3},-\dfrac{2}{3}\right)\)
- \(\left(\dfrac{2}{3\sqrt{5}},-\dfrac{4}{3\sqrt{5}},\dfrac{5}{3\sqrt{5}}\right)\)
- \(\left(\dfrac{-4}{\sqrt{57}},\dfrac{4}{\sqrt{57}},\dfrac{-5}{\sqrt{57}}\right)\)
- \(\left(\dfrac{2}{\sqrt{38}},-\dfrac{3}{\sqrt{38}},\dfrac{5}{\sqrt{38}}\right)\)
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Write the equation of the plane in intercept form.
If a plane cuts intercepts \(a,b,c\) on the \(X,Y,Z\)-axes respectively, then its equation is \[ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 \] Here, \[ a=-2,\quad b=\frac{4}{3},\quad c=-\frac{4}{5} \] Therefore, \[ \frac{x}{-2}+\frac{y}{\frac{4}{3}}+\frac{z}{-\frac{4}{5}}=1 \]
Step 2: Simplify the plane equation.
We get \[ -\frac{x}{2}+\frac{3y}{4}-\frac{5z}{4}=1 \] Multiplying by \(4\), \[ -2x+3y-5z=4 \] Equivalently, \[ 2x-3y+5z+4=0 \]
Step 3: Identify the normal vector.
For a plane \[ Ax+By+Cz+D=0, \] a normal vector is \[ (A,B,C) \] Hence, for \[ 2x-3y+5z+4=0, \] a normal vector is \[ (2,-3,5) \]
Step 4: Find the direction cosines of the normal.
The magnitude of the normal vector is \[ \sqrt{2^2+(-3)^2+5^2} \] \[ =\sqrt{4+9+25} \] \[ =\sqrt{38} \] Therefore, the direction cosines are \[ \left(\frac{2}{\sqrt{38}},\frac{-3}{\sqrt{38}},\frac{5}{\sqrt{38}}\right) \]
Step 5: Final conclusion.
Hence, the required direction cosines are \[ \boxed{\left(\frac{2}{\sqrt{38}},-\frac{3}{\sqrt{38}},\frac{5}{\sqrt{38}}\right)} \]
If a plane cuts intercepts \(a,b,c\) on the \(X,Y,Z\)-axes respectively, then its equation is \[ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 \] Here, \[ a=-2,\quad b=\frac{4}{3},\quad c=-\frac{4}{5} \] Therefore, \[ \frac{x}{-2}+\frac{y}{\frac{4}{3}}+\frac{z}{-\frac{4}{5}}=1 \]
Step 2: Simplify the plane equation.
We get \[ -\frac{x}{2}+\frac{3y}{4}-\frac{5z}{4}=1 \] Multiplying by \(4\), \[ -2x+3y-5z=4 \] Equivalently, \[ 2x-3y+5z+4=0 \]
Step 3: Identify the normal vector.
For a plane \[ Ax+By+Cz+D=0, \] a normal vector is \[ (A,B,C) \] Hence, for \[ 2x-3y+5z+4=0, \] a normal vector is \[ (2,-3,5) \]
Step 4: Find the direction cosines of the normal.
The magnitude of the normal vector is \[ \sqrt{2^2+(-3)^2+5^2} \] \[ =\sqrt{4+9+25} \] \[ =\sqrt{38} \] Therefore, the direction cosines are \[ \left(\frac{2}{\sqrt{38}},\frac{-3}{\sqrt{38}},\frac{5}{\sqrt{38}}\right) \]
Step 5: Final conclusion.
Hence, the required direction cosines are \[ \boxed{\left(\frac{2}{\sqrt{38}},-\frac{3}{\sqrt{38}},\frac{5}{\sqrt{38}}\right)} \]
Was this answer helpful?
0
0
Top AP EAPCET Mathematics Questions
- $D = \left\{x\in\mathbb{R}: f\left(x\right) =\sqrt{\frac{x - \left|x\right|}{x - \left[x\right]}} \text{is defined} \right\}$ and $C$ be the range of the real function $g(x) = \frac{2x}{4 + x^{2}}$. Then $D \cap C$
- $f\left(x\right) = \frac{x}{e^{x}-1} + \frac{x}{2} + 2 \cos^{3} \frac{x}{2} $ on $R-\left\{0\right\} $ is
- Consider the following lists.
List-I List-II (A) $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ (I) $[\frac{1}{3} , 1 ]$ (B) $(x)=|[x]|,x \in [R$ (II) Z (C) $h(x) = |x - [x]| , x \in [R$ (III) W (D) $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ (IV) [0, 1) (V) { -1, 1} - If $[x]$ is the greatest integer less than or equal to $x$ and $|x|$ is the modulus of $x$. then the system of three equations $2x + 3 | y | + 5[z] = 0, x + |y| - 2[z] = 4, x + |y| + |z| = 1$ has
- Investigate the values of $\lambda$ and $\mu$ for the system $x+2 y+3 z=6, x+3 y+5 z=9$, $2 x+5 y+\lambda z=\mu$ and match the values in List - I with the items in List - II.
List I List II (A) $\lambda=8, \mu \neq 15$ 1. Infinitely many solutions (B) $\lambda \neq 8, \mu \in R$ 2. No solution (C) $\lambda=8, \mu=15$ 3. Unique solution
View More Questions
Top AP EAPCET Plane Questions
- Let \(\pi\) be the plane passing through the point \((3,-3,1)\) and perpendicular to the line joining the points \[ (3,4,-1) \] and \[ (2,-1,5). \] If the equation of the plane containing the points \[ (3,4,-1),\quad (-1,2,5) \] and perpendicular to the plane \(\pi\) is \[ ax+y+cz-d=0, \] then \[ 3(a+c)= \]
- If the equation of the plane passing through the point \(A(-2,1,3)\) and perpendicular to the vector \(3\vec{i}+\vec{j}+5\vec{k}\) is \(ax+by+cz+d=0\), then \(\dfrac{a+b}{c+d}=\)
- If the equation of the plane which is at a distance of \(\dfrac{1}{3}\) units from the origin and perpendicular to a line whose directional ratios are \((1,2,2)\) is \[ x+py+qz+r=0 \] then \[ \sqrt{p^2+q^2+r^2}= \]
- Let the plane \(\pi\) pass through the point \((1,0,1)\) and perpendicular to the planes \[ 2x+3y-z=2 \] and \[ x-y+2z=1. \] Let the equation of the plane passing through the point \((11,7,5)\) and parallel to the plane \(\pi\) be \[ ax+by-z+d=0. \] Then \[ \frac{a}{b}+\frac{b}{d}= \]
- If \(a,b,c\) are the intercepts made by the plane passing through the point \((1,2,3)\) parallel to the plane \(3x+4y-5z=0\) on \(X,Y,Z\)-axes respectively, then \(3a+b+5c=\)
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions