121390
The sine of the angle between the straight line \(\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\) and the plane \(2 \mathrm{x}-2 \mathrm{y}+\mathrm{z}=5\) is
1 \(\frac{2 \sqrt{3}}{5}\)
2 \(\frac{\sqrt{2}}{10}\)
3 \(\frac{4}{5 \sqrt{2}}\)
4 \(\frac{\sqrt{5}}{6}\)
Explanation:
B Given, Line : \(\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\)
Plane : \(2 \mathrm{x}-2 \mathrm{y}+\mathrm{z}=5\)
Angle between line and plane -
\(\sin \theta=\left \vert\frac{ \vert\mathrm{al}+\mathrm{bm}+\mathrm{cn} \vert}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2} \cdot \sqrt{\mathrm{l}^2+\mathrm{m}^2+\mathrm{n}^2}}\right \vert\)
\(\Rightarrow \sin \theta=\left(\frac{3 \times 2+4 \times(-2)+5 \times 1}{\sqrt{9+16+25} \cdot \sqrt{4+4+1}}\right)\)
\(\Rightarrow \sin \theta=\frac{3}{15 \sqrt{2}}\)
\(\Rightarrow \quad \sin \theta=\frac{\sqrt{2}}{10}\)
WB JEE-2020
Three Dimensional Geometry
121391
If the equation of the plane bisection the line segment joining the points \(P(3,2,4)\) and \(Q(-\) \(1,0,-2)\) and perpendicular to \(P Q\) is \(\mathbf{a x}+\) by \(c z\) \(+d=0\), then ac + bd
1 0
2 12
3 6
4 1
Explanation:
A Given, \(\quad\)\(P(3,2,4)\) and \(Q(-1,0,-2)\) \(\Rightarrow \overrightarrow{\mathrm{PQ}}=-4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}\) Midpoint of \(\overrightarrow{\mathrm{PQ}}\) is \(\mathrm{M}=\left(\frac{3-1}{2}, \frac{2+0}{2}, \frac{4-2}{2}\right)\) \(\Rightarrow \mathrm{M}=(1,1,1)\) Equation of required plane \(-4(x-1)-2(y-1)-6(z-1)=0\) \(\Rightarrow-4 \mathrm{x}-2 \mathrm{y}-6 \mathrm{z}=-12\) \(\text { or } 2 x+y+3 z=6\) \(\text { Here, } a=2, b=1, c=3 \text {, and } d=-6\) \(\text { Then, }\) \(\mathrm{ac}+\mathrm{bd}=2 \times 3+1 \times(-6)\) \(=6-6\) \(a c+b d=0\)
AP EAMCET-23.04.2018
Three Dimensional Geometry
121393
Foot of the perpendicular drawn from the point \((1,3,4)\) to the plane \(2 x-y+z+3=0\) is
1 \((-1,4,3)\)
2 \((1,2,-3)\)
3 \((0,-4,-7)\)
4 \((-3,5,2)\)
Explanation:
A Given, Point \(\mathrm{A}=(1,3,4)\) Plane : \(2 \mathrm{x}-\mathrm{y}+\mathrm{z}+3=0\) Equation of line passing point \(\mathrm{A}\) and parallel to normal of the plane is \(\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-3}{-1}=\frac{\mathrm{z}-4}{1}=\mathrm{k}\) Any point on this line \(\mathrm{B}=(2 \mathrm{k}+1,-\mathrm{k}+3, \mathrm{k}+4)\) Now, this point also lie on given plane. Hence, \(2(2 \mathrm{k}+1)-(-\mathrm{k}+3)+(\mathrm{k}+4)+3=0\) \(6 \mathrm{k}+6=0\) \(\mathrm{k}=-1\) Therefore, foot of perpendicular is \(\mathrm{B}=(-1,4,3)\)
Karnataka CET-2019
Three Dimensional Geometry
121394
If \((-4,5)\) is the image of the point \((6,1)\) with respect to the line \(L\), then \(L\) is given by
1 \(5 x+2 y=1\)
2 \(5 \mathrm{x}-2 \mathrm{y}=0\)
3 \(5 \mathrm{x}-2 \mathrm{y}+1=0\)
4 \(2 x-5 y+1=0\)
Explanation:
C Point \(R=\left(\frac{-4+6}{2}, \frac{5+1}{2}\right)\) \(=(1,3)\) Slope of \(P Q=\frac{5-1}{-4-6}=\frac{4}{-10}=\frac{-2}{5}\) Since, Line (L) is perpendicular to PQ Slope of Line \((\mathrm{L})=\frac{5}{2}\) So, the equation of line (L) passing through \(\mathrm{R}(1,3)\) and having slope \(\frac{5}{2}\) is given by \(y-3=\frac{5}{2}(x-1)\) \(\Rightarrow 2 y-6=5 x-5\) \(\Rightarrow 5 x-2 y+1=0\)
121390
The sine of the angle between the straight line \(\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\) and the plane \(2 \mathrm{x}-2 \mathrm{y}+\mathrm{z}=5\) is
1 \(\frac{2 \sqrt{3}}{5}\)
2 \(\frac{\sqrt{2}}{10}\)
3 \(\frac{4}{5 \sqrt{2}}\)
4 \(\frac{\sqrt{5}}{6}\)
Explanation:
B Given, Line : \(\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\)
Plane : \(2 \mathrm{x}-2 \mathrm{y}+\mathrm{z}=5\)
Angle between line and plane -
\(\sin \theta=\left \vert\frac{ \vert\mathrm{al}+\mathrm{bm}+\mathrm{cn} \vert}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2} \cdot \sqrt{\mathrm{l}^2+\mathrm{m}^2+\mathrm{n}^2}}\right \vert\)
\(\Rightarrow \sin \theta=\left(\frac{3 \times 2+4 \times(-2)+5 \times 1}{\sqrt{9+16+25} \cdot \sqrt{4+4+1}}\right)\)
\(\Rightarrow \sin \theta=\frac{3}{15 \sqrt{2}}\)
\(\Rightarrow \quad \sin \theta=\frac{\sqrt{2}}{10}\)
WB JEE-2020
Three Dimensional Geometry
121391
If the equation of the plane bisection the line segment joining the points \(P(3,2,4)\) and \(Q(-\) \(1,0,-2)\) and perpendicular to \(P Q\) is \(\mathbf{a x}+\) by \(c z\) \(+d=0\), then ac + bd
1 0
2 12
3 6
4 1
Explanation:
A Given, \(\quad\)\(P(3,2,4)\) and \(Q(-1,0,-2)\) \(\Rightarrow \overrightarrow{\mathrm{PQ}}=-4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}\) Midpoint of \(\overrightarrow{\mathrm{PQ}}\) is \(\mathrm{M}=\left(\frac{3-1}{2}, \frac{2+0}{2}, \frac{4-2}{2}\right)\) \(\Rightarrow \mathrm{M}=(1,1,1)\) Equation of required plane \(-4(x-1)-2(y-1)-6(z-1)=0\) \(\Rightarrow-4 \mathrm{x}-2 \mathrm{y}-6 \mathrm{z}=-12\) \(\text { or } 2 x+y+3 z=6\) \(\text { Here, } a=2, b=1, c=3 \text {, and } d=-6\) \(\text { Then, }\) \(\mathrm{ac}+\mathrm{bd}=2 \times 3+1 \times(-6)\) \(=6-6\) \(a c+b d=0\)
AP EAMCET-23.04.2018
Three Dimensional Geometry
121393
Foot of the perpendicular drawn from the point \((1,3,4)\) to the plane \(2 x-y+z+3=0\) is
1 \((-1,4,3)\)
2 \((1,2,-3)\)
3 \((0,-4,-7)\)
4 \((-3,5,2)\)
Explanation:
A Given, Point \(\mathrm{A}=(1,3,4)\) Plane : \(2 \mathrm{x}-\mathrm{y}+\mathrm{z}+3=0\) Equation of line passing point \(\mathrm{A}\) and parallel to normal of the plane is \(\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-3}{-1}=\frac{\mathrm{z}-4}{1}=\mathrm{k}\) Any point on this line \(\mathrm{B}=(2 \mathrm{k}+1,-\mathrm{k}+3, \mathrm{k}+4)\) Now, this point also lie on given plane. Hence, \(2(2 \mathrm{k}+1)-(-\mathrm{k}+3)+(\mathrm{k}+4)+3=0\) \(6 \mathrm{k}+6=0\) \(\mathrm{k}=-1\) Therefore, foot of perpendicular is \(\mathrm{B}=(-1,4,3)\)
Karnataka CET-2019
Three Dimensional Geometry
121394
If \((-4,5)\) is the image of the point \((6,1)\) with respect to the line \(L\), then \(L\) is given by
1 \(5 x+2 y=1\)
2 \(5 \mathrm{x}-2 \mathrm{y}=0\)
3 \(5 \mathrm{x}-2 \mathrm{y}+1=0\)
4 \(2 x-5 y+1=0\)
Explanation:
C Point \(R=\left(\frac{-4+6}{2}, \frac{5+1}{2}\right)\) \(=(1,3)\) Slope of \(P Q=\frac{5-1}{-4-6}=\frac{4}{-10}=\frac{-2}{5}\) Since, Line (L) is perpendicular to PQ Slope of Line \((\mathrm{L})=\frac{5}{2}\) So, the equation of line (L) passing through \(\mathrm{R}(1,3)\) and having slope \(\frac{5}{2}\) is given by \(y-3=\frac{5}{2}(x-1)\) \(\Rightarrow 2 y-6=5 x-5\) \(\Rightarrow 5 x-2 y+1=0\)
121390
The sine of the angle between the straight line \(\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\) and the plane \(2 \mathrm{x}-2 \mathrm{y}+\mathrm{z}=5\) is
1 \(\frac{2 \sqrt{3}}{5}\)
2 \(\frac{\sqrt{2}}{10}\)
3 \(\frac{4}{5 \sqrt{2}}\)
4 \(\frac{\sqrt{5}}{6}\)
Explanation:
B Given, Line : \(\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\)
Plane : \(2 \mathrm{x}-2 \mathrm{y}+\mathrm{z}=5\)
Angle between line and plane -
\(\sin \theta=\left \vert\frac{ \vert\mathrm{al}+\mathrm{bm}+\mathrm{cn} \vert}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2} \cdot \sqrt{\mathrm{l}^2+\mathrm{m}^2+\mathrm{n}^2}}\right \vert\)
\(\Rightarrow \sin \theta=\left(\frac{3 \times 2+4 \times(-2)+5 \times 1}{\sqrt{9+16+25} \cdot \sqrt{4+4+1}}\right)\)
\(\Rightarrow \sin \theta=\frac{3}{15 \sqrt{2}}\)
\(\Rightarrow \quad \sin \theta=\frac{\sqrt{2}}{10}\)
WB JEE-2020
Three Dimensional Geometry
121391
If the equation of the plane bisection the line segment joining the points \(P(3,2,4)\) and \(Q(-\) \(1,0,-2)\) and perpendicular to \(P Q\) is \(\mathbf{a x}+\) by \(c z\) \(+d=0\), then ac + bd
1 0
2 12
3 6
4 1
Explanation:
A Given, \(\quad\)\(P(3,2,4)\) and \(Q(-1,0,-2)\) \(\Rightarrow \overrightarrow{\mathrm{PQ}}=-4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}\) Midpoint of \(\overrightarrow{\mathrm{PQ}}\) is \(\mathrm{M}=\left(\frac{3-1}{2}, \frac{2+0}{2}, \frac{4-2}{2}\right)\) \(\Rightarrow \mathrm{M}=(1,1,1)\) Equation of required plane \(-4(x-1)-2(y-1)-6(z-1)=0\) \(\Rightarrow-4 \mathrm{x}-2 \mathrm{y}-6 \mathrm{z}=-12\) \(\text { or } 2 x+y+3 z=6\) \(\text { Here, } a=2, b=1, c=3 \text {, and } d=-6\) \(\text { Then, }\) \(\mathrm{ac}+\mathrm{bd}=2 \times 3+1 \times(-6)\) \(=6-6\) \(a c+b d=0\)
AP EAMCET-23.04.2018
Three Dimensional Geometry
121393
Foot of the perpendicular drawn from the point \((1,3,4)\) to the plane \(2 x-y+z+3=0\) is
1 \((-1,4,3)\)
2 \((1,2,-3)\)
3 \((0,-4,-7)\)
4 \((-3,5,2)\)
Explanation:
A Given, Point \(\mathrm{A}=(1,3,4)\) Plane : \(2 \mathrm{x}-\mathrm{y}+\mathrm{z}+3=0\) Equation of line passing point \(\mathrm{A}\) and parallel to normal of the plane is \(\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-3}{-1}=\frac{\mathrm{z}-4}{1}=\mathrm{k}\) Any point on this line \(\mathrm{B}=(2 \mathrm{k}+1,-\mathrm{k}+3, \mathrm{k}+4)\) Now, this point also lie on given plane. Hence, \(2(2 \mathrm{k}+1)-(-\mathrm{k}+3)+(\mathrm{k}+4)+3=0\) \(6 \mathrm{k}+6=0\) \(\mathrm{k}=-1\) Therefore, foot of perpendicular is \(\mathrm{B}=(-1,4,3)\)
Karnataka CET-2019
Three Dimensional Geometry
121394
If \((-4,5)\) is the image of the point \((6,1)\) with respect to the line \(L\), then \(L\) is given by
1 \(5 x+2 y=1\)
2 \(5 \mathrm{x}-2 \mathrm{y}=0\)
3 \(5 \mathrm{x}-2 \mathrm{y}+1=0\)
4 \(2 x-5 y+1=0\)
Explanation:
C Point \(R=\left(\frac{-4+6}{2}, \frac{5+1}{2}\right)\) \(=(1,3)\) Slope of \(P Q=\frac{5-1}{-4-6}=\frac{4}{-10}=\frac{-2}{5}\) Since, Line (L) is perpendicular to PQ Slope of Line \((\mathrm{L})=\frac{5}{2}\) So, the equation of line (L) passing through \(\mathrm{R}(1,3)\) and having slope \(\frac{5}{2}\) is given by \(y-3=\frac{5}{2}(x-1)\) \(\Rightarrow 2 y-6=5 x-5\) \(\Rightarrow 5 x-2 y+1=0\)
121390
The sine of the angle between the straight line \(\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\) and the plane \(2 \mathrm{x}-2 \mathrm{y}+\mathrm{z}=5\) is
1 \(\frac{2 \sqrt{3}}{5}\)
2 \(\frac{\sqrt{2}}{10}\)
3 \(\frac{4}{5 \sqrt{2}}\)
4 \(\frac{\sqrt{5}}{6}\)
Explanation:
B Given, Line : \(\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\)
Plane : \(2 \mathrm{x}-2 \mathrm{y}+\mathrm{z}=5\)
Angle between line and plane -
\(\sin \theta=\left \vert\frac{ \vert\mathrm{al}+\mathrm{bm}+\mathrm{cn} \vert}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2} \cdot \sqrt{\mathrm{l}^2+\mathrm{m}^2+\mathrm{n}^2}}\right \vert\)
\(\Rightarrow \sin \theta=\left(\frac{3 \times 2+4 \times(-2)+5 \times 1}{\sqrt{9+16+25} \cdot \sqrt{4+4+1}}\right)\)
\(\Rightarrow \sin \theta=\frac{3}{15 \sqrt{2}}\)
\(\Rightarrow \quad \sin \theta=\frac{\sqrt{2}}{10}\)
WB JEE-2020
Three Dimensional Geometry
121391
If the equation of the plane bisection the line segment joining the points \(P(3,2,4)\) and \(Q(-\) \(1,0,-2)\) and perpendicular to \(P Q\) is \(\mathbf{a x}+\) by \(c z\) \(+d=0\), then ac + bd
1 0
2 12
3 6
4 1
Explanation:
A Given, \(\quad\)\(P(3,2,4)\) and \(Q(-1,0,-2)\) \(\Rightarrow \overrightarrow{\mathrm{PQ}}=-4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}\) Midpoint of \(\overrightarrow{\mathrm{PQ}}\) is \(\mathrm{M}=\left(\frac{3-1}{2}, \frac{2+0}{2}, \frac{4-2}{2}\right)\) \(\Rightarrow \mathrm{M}=(1,1,1)\) Equation of required plane \(-4(x-1)-2(y-1)-6(z-1)=0\) \(\Rightarrow-4 \mathrm{x}-2 \mathrm{y}-6 \mathrm{z}=-12\) \(\text { or } 2 x+y+3 z=6\) \(\text { Here, } a=2, b=1, c=3 \text {, and } d=-6\) \(\text { Then, }\) \(\mathrm{ac}+\mathrm{bd}=2 \times 3+1 \times(-6)\) \(=6-6\) \(a c+b d=0\)
AP EAMCET-23.04.2018
Three Dimensional Geometry
121393
Foot of the perpendicular drawn from the point \((1,3,4)\) to the plane \(2 x-y+z+3=0\) is
1 \((-1,4,3)\)
2 \((1,2,-3)\)
3 \((0,-4,-7)\)
4 \((-3,5,2)\)
Explanation:
A Given, Point \(\mathrm{A}=(1,3,4)\) Plane : \(2 \mathrm{x}-\mathrm{y}+\mathrm{z}+3=0\) Equation of line passing point \(\mathrm{A}\) and parallel to normal of the plane is \(\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-3}{-1}=\frac{\mathrm{z}-4}{1}=\mathrm{k}\) Any point on this line \(\mathrm{B}=(2 \mathrm{k}+1,-\mathrm{k}+3, \mathrm{k}+4)\) Now, this point also lie on given plane. Hence, \(2(2 \mathrm{k}+1)-(-\mathrm{k}+3)+(\mathrm{k}+4)+3=0\) \(6 \mathrm{k}+6=0\) \(\mathrm{k}=-1\) Therefore, foot of perpendicular is \(\mathrm{B}=(-1,4,3)\)
Karnataka CET-2019
Three Dimensional Geometry
121394
If \((-4,5)\) is the image of the point \((6,1)\) with respect to the line \(L\), then \(L\) is given by
1 \(5 x+2 y=1\)
2 \(5 \mathrm{x}-2 \mathrm{y}=0\)
3 \(5 \mathrm{x}-2 \mathrm{y}+1=0\)
4 \(2 x-5 y+1=0\)
Explanation:
C Point \(R=\left(\frac{-4+6}{2}, \frac{5+1}{2}\right)\) \(=(1,3)\) Slope of \(P Q=\frac{5-1}{-4-6}=\frac{4}{-10}=\frac{-2}{5}\) Since, Line (L) is perpendicular to PQ Slope of Line \((\mathrm{L})=\frac{5}{2}\) So, the equation of line (L) passing through \(\mathrm{R}(1,3)\) and having slope \(\frac{5}{2}\) is given by \(y-3=\frac{5}{2}(x-1)\) \(\Rightarrow 2 y-6=5 x-5\) \(\Rightarrow 5 x-2 y+1=0\)
