121439
A straight line \(\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{a}}+\lambda \overrightarrow{\mathrm{b}}\) meets the plane \(\overrightarrow{\mathbf{r}} \cdot \hat{\mathbf{n}}=p\) in the point \(P\) whose position vector is :
B Since straight line \(\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}\) \(\text { Meets the plane } \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{n}}=\mathrm{p} \text { at point } \mathrm{p}\) \(\text { then }(\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{n}}=\mathrm{p}\) \(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{n}}+\lambda(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{n}})=\mathrm{p}\) \(\Rightarrow \lambda=\frac{\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{n}}}{\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{n}}}\) \(\text { Substituting value of } \lambda \text { in (i), }\) \(\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\left(\frac{\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{n}}}{\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{n}}}\right) \overrightarrow{\mathrm{b}}\) Hence, position vector of point \(\mathrm{P}\) is \(r=\overrightarrow{\mathrm{a}}+\left(\frac{\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{n}}}{\overrightarrow{\mathrm{b}} \cdot \vec{n}}\right) \overrightarrow{\mathrm{b}}\) Meets the plane \(\vec{r} . \vec{n}=p\) at point \(p\) then \((\vec{a}+\lambda \vec{b}) \cdot \vec{n}=p\) Substituting value of \(\lambda\) in (i),
AMU-2014
Three Dimensional Geometry
121445
If the \(2 x-5 y+z=8\) and \(2 \lambda x-15 y+\lambda z+6=0\) are parallel to each other, then value of \(\lambda\) is
1 \(\frac{1}{3}\)
2 -3
3 3
4 2
Explanation:
C Given planes are parallel. Therefore, normal vector to their plane is also parallel. \(\therefore \frac{2}{2 \lambda}=\frac{-5}{-15} \Rightarrow \frac{1}{\lambda}=\frac{1}{3} \Rightarrow \lambda=3\)
MHT CET-2020
Three Dimensional Geometry
121426
The foot of the perpendicular from a point on the circle \(x^2+y^2=1, z=0\) to the plane \(2 x+3 y\) \(+z=6\) lies on which one of the following curves?
121439
A straight line \(\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{a}}+\lambda \overrightarrow{\mathrm{b}}\) meets the plane \(\overrightarrow{\mathbf{r}} \cdot \hat{\mathbf{n}}=p\) in the point \(P\) whose position vector is :
B Since straight line \(\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}\) \(\text { Meets the plane } \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{n}}=\mathrm{p} \text { at point } \mathrm{p}\) \(\text { then }(\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{n}}=\mathrm{p}\) \(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{n}}+\lambda(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{n}})=\mathrm{p}\) \(\Rightarrow \lambda=\frac{\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{n}}}{\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{n}}}\) \(\text { Substituting value of } \lambda \text { in (i), }\) \(\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\left(\frac{\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{n}}}{\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{n}}}\right) \overrightarrow{\mathrm{b}}\) Hence, position vector of point \(\mathrm{P}\) is \(r=\overrightarrow{\mathrm{a}}+\left(\frac{\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{n}}}{\overrightarrow{\mathrm{b}} \cdot \vec{n}}\right) \overrightarrow{\mathrm{b}}\) Meets the plane \(\vec{r} . \vec{n}=p\) at point \(p\) then \((\vec{a}+\lambda \vec{b}) \cdot \vec{n}=p\) Substituting value of \(\lambda\) in (i),
AMU-2014
Three Dimensional Geometry
121445
If the \(2 x-5 y+z=8\) and \(2 \lambda x-15 y+\lambda z+6=0\) are parallel to each other, then value of \(\lambda\) is
1 \(\frac{1}{3}\)
2 -3
3 3
4 2
Explanation:
C Given planes are parallel. Therefore, normal vector to their plane is also parallel. \(\therefore \frac{2}{2 \lambda}=\frac{-5}{-15} \Rightarrow \frac{1}{\lambda}=\frac{1}{3} \Rightarrow \lambda=3\)
MHT CET-2020
Three Dimensional Geometry
121426
The foot of the perpendicular from a point on the circle \(x^2+y^2=1, z=0\) to the plane \(2 x+3 y\) \(+z=6\) lies on which one of the following curves?
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Three Dimensional Geometry
121439
A straight line \(\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{a}}+\lambda \overrightarrow{\mathrm{b}}\) meets the plane \(\overrightarrow{\mathbf{r}} \cdot \hat{\mathbf{n}}=p\) in the point \(P\) whose position vector is :
B Since straight line \(\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}\) \(\text { Meets the plane } \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{n}}=\mathrm{p} \text { at point } \mathrm{p}\) \(\text { then }(\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{n}}=\mathrm{p}\) \(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{n}}+\lambda(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{n}})=\mathrm{p}\) \(\Rightarrow \lambda=\frac{\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{n}}}{\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{n}}}\) \(\text { Substituting value of } \lambda \text { in (i), }\) \(\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\left(\frac{\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{n}}}{\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{n}}}\right) \overrightarrow{\mathrm{b}}\) Hence, position vector of point \(\mathrm{P}\) is \(r=\overrightarrow{\mathrm{a}}+\left(\frac{\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{n}}}{\overrightarrow{\mathrm{b}} \cdot \vec{n}}\right) \overrightarrow{\mathrm{b}}\) Meets the plane \(\vec{r} . \vec{n}=p\) at point \(p\) then \((\vec{a}+\lambda \vec{b}) \cdot \vec{n}=p\) Substituting value of \(\lambda\) in (i),
AMU-2014
Three Dimensional Geometry
121445
If the \(2 x-5 y+z=8\) and \(2 \lambda x-15 y+\lambda z+6=0\) are parallel to each other, then value of \(\lambda\) is
1 \(\frac{1}{3}\)
2 -3
3 3
4 2
Explanation:
C Given planes are parallel. Therefore, normal vector to their plane is also parallel. \(\therefore \frac{2}{2 \lambda}=\frac{-5}{-15} \Rightarrow \frac{1}{\lambda}=\frac{1}{3} \Rightarrow \lambda=3\)
MHT CET-2020
Three Dimensional Geometry
121426
The foot of the perpendicular from a point on the circle \(x^2+y^2=1, z=0\) to the plane \(2 x+3 y\) \(+z=6\) lies on which one of the following curves?