121256
If the line passing through \(A(\lambda, 2,3)\) and \(B(1,2, \mu)\) is parallel to the line \(\mathbf{x}-\mathbf{1}=\mathbf{y}-\mathbf{2}=\mathbf{z}-\mathbf{3}\), then \(\lambda+\mu=\)
1 1
2 4
3 2
4 3
Explanation:
B The equation of line passing through \(\mathrm{A}(\lambda, 2,3)\) and \(\mathrm{B}(1,2, \mu)\) is \(\frac{x-\lambda}{\lambda-1}=\frac{y-2}{2-2}=\frac{z-3}{3-\mu}\) i.e. \(\frac{x-\lambda}{\lambda-1}=\frac{y-2}{0}=\frac{z-3}{3-\mu}\)... The line (i) is parallel to the line \(\mathrm{x}-1=\mathrm{y}-2=\mathrm{z}-3\) \(\therefore \lambda-1=0=3-\mu \Rightarrow \lambda=1\) and \(\mu=3\) Hence, \(\lambda+\mu=1+3=4\)
MHT CET-2019
Three Dimensional Geometry
121257
Equation of planes parallel to the plane \(x-2 y+2 z+4=0\) which are at a distance of one unit form the point \((1,2,3)\) are
1 \(x+2 y+2 z=6, x+2 y+2 z=0\)
2 \(x-2 y+2 z=0, x-2 y+2 z-6=0\)
3 \(x-2 y-6=0, x-2 y+z=6\)
4 \(x+2 y+2 z=-6, x+2 y+2 z=5\)
Explanation:
B The equation of planes parallel to the plane \(\mathrm{x}-2 \mathrm{y}+2 \mathrm{z}+4=0\) is \(\mathrm{x}-2 \mathrm{y}+2 \mathrm{z}+\lambda=0\)
The required planes are at a distance of one unit from the point \((1,2,3)\).
\(\left|\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{cz}_1+\mathrm{d}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}\right|=1\)
\(\left|\frac{1(1)+(-2)(2)+2(3)+\lambda}{\sqrt{1+4+4}}\right|=1\)
\(\left|\frac{1-4+6+\lambda}{3}\right|=1\)
\(3+\lambda= \pm 3\)
\(\therefore 3+\lambda=3 \quad\) or \(\quad 3+\lambda=-3\)
\(\lambda=0 \quad\) or \(\quad \lambda=-6\)
Hence, the equation of planes are \(x-2 y+2 z=0\) and \(x-2 y+2 z-6=0\)
MHT CET-2019
Three Dimensional Geometry
121258
If the foot of the perpendicular drawn form the point \((0,0,0)\) to the plane is \((4,-2,-5)\), then the equation of the plane is
1 \(4 x-2 y+5 z=-5\)
2 \(4 x-2 y-5 z=45\)
3 \(4 x+2 y-5 z=37\)
4 \(4 x+2 y+5 z=-13\)
Explanation:
B The required plane passes through the point \(\mathrm{M}(4,-2,-5)\) and is perpendicular to \(\overline{\mathrm{OM}}\) \(\therefore \overline{\mathrm{a}}=4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}\) and \(\overline{\mathrm{OM}}=\overline{\mathrm{n}}=4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}\), which is normal to the plane. \(\therefore \overline{\mathrm{a}}=\overline{\mathrm{n}}\) \(\therefore\) The equation of the plane is \(\overline{\mathrm{r}} \cdot \overline{\mathrm{n}}=\overline{\mathrm{a}} \cdot \overline{\mathrm{n}}\) \(\therefore \overline{\mathrm{r}} \cdot(4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \mathrm{k})=(4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}) \cdot(4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})\) \(\quad=16+4+25\) \(\overline{\mathrm{r}} \cdot(4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})=45\) \(\text { Taking } \overline{\mathrm{r}}=x \hat{\mathrm{i}}+\mathrm{y}+\mathrm{j} \hat{\mathrm{k}}\) \((x \hat{\mathrm{i}}+\mathrm{y}+\mathrm{j} \hat{\mathrm{k}}) \cdot(4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})=45 \Rightarrow 4 \mathrm{x}-2 \mathrm{y}-5 \mathrm{z}=45\)Taking \(\overline{\mathrm{r}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}\)
MHT CET-2019
Three Dimensional Geometry
121260
If lines \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\) and \(x-3=\frac{y-k}{2}=z\) interest, then the value of \(k\) is
1 \(\frac{9}{2}\)
2 \(\frac{1}{2}\)
3 \(\frac{5}{2}\)
4 \(\frac{7}{2}\)
Explanation:
A Points on the line are \((1,-1,1)\) and \((3, \mathrm{k}, 0)\) and direction ratios or line are 2,3,4 and 1,2,1.
Since lines intersect, then lies are coplanar,
\(\therefore\left|\begin{array}{ccc}\mathrm{x}_2-\mathrm{x}_1 & \mathrm{y}_2-\mathrm{y}_1 & \mathrm{z}_2-\mathrm{z}_1 \\ \mathrm{a}_1 & \mathrm{~b}_1 & \mathrm{c}_1 \\ \mathrm{a}_2 & \mathrm{~b}_2 & \mathrm{c}_2\end{array}\right|=0 \Rightarrow\left|\begin{array}{ccc}2 & \mathrm{k}+1 & -1 \\ 2 & 3 & 4 \\ 1 & 2 & 1\end{array}\right|=0\)
\(\therefore 2(-5)-(\mathrm{k}+1)(-2)-1(1)=0 \Rightarrow-11+2 \mathrm{k}+2=0 \Rightarrow \mathrm{k}=\frac{9}{2}\)
MHT CET-2018
Three Dimensional Geometry
121261
If planes \(x-c y-b y=0, c x-y+a z=0\) and \(b x\) \(+a y-z=0\) pass through a straight line, then \(\mathbf{a}^2+\mathbf{b}^2+\mathbf{c}^2=\)
1 \(1-\mathrm{abc}\)
2 abc -1
3 1-2abc
4 \(2 \mathrm{abc}-1\)
Explanation:
C Given, Planes are -
\(x-c y-b z=0\)
\(c x-y+a z=0\)
\(b x+a y-z=0\)
\(\therefore\) Planes are concurrent
\(\left|\begin{array}{ccc}1 & -c & -b \\ c & -1 & a \\ b & a & -1\end{array}\right|=0\)
\(1\left(1-a^2\right)+c(-c-a b)-b(a c+b)=0\)
\(1-a^2-c^2-a b c-a b c-b^2=0 \Rightarrow a^2+b^2+c^2=1-2 a b c\)
121256
If the line passing through \(A(\lambda, 2,3)\) and \(B(1,2, \mu)\) is parallel to the line \(\mathbf{x}-\mathbf{1}=\mathbf{y}-\mathbf{2}=\mathbf{z}-\mathbf{3}\), then \(\lambda+\mu=\)
1 1
2 4
3 2
4 3
Explanation:
B The equation of line passing through \(\mathrm{A}(\lambda, 2,3)\) and \(\mathrm{B}(1,2, \mu)\) is \(\frac{x-\lambda}{\lambda-1}=\frac{y-2}{2-2}=\frac{z-3}{3-\mu}\) i.e. \(\frac{x-\lambda}{\lambda-1}=\frac{y-2}{0}=\frac{z-3}{3-\mu}\)... The line (i) is parallel to the line \(\mathrm{x}-1=\mathrm{y}-2=\mathrm{z}-3\) \(\therefore \lambda-1=0=3-\mu \Rightarrow \lambda=1\) and \(\mu=3\) Hence, \(\lambda+\mu=1+3=4\)
MHT CET-2019
Three Dimensional Geometry
121257
Equation of planes parallel to the plane \(x-2 y+2 z+4=0\) which are at a distance of one unit form the point \((1,2,3)\) are
1 \(x+2 y+2 z=6, x+2 y+2 z=0\)
2 \(x-2 y+2 z=0, x-2 y+2 z-6=0\)
3 \(x-2 y-6=0, x-2 y+z=6\)
4 \(x+2 y+2 z=-6, x+2 y+2 z=5\)
Explanation:
B The equation of planes parallel to the plane \(\mathrm{x}-2 \mathrm{y}+2 \mathrm{z}+4=0\) is \(\mathrm{x}-2 \mathrm{y}+2 \mathrm{z}+\lambda=0\)
The required planes are at a distance of one unit from the point \((1,2,3)\).
\(\left|\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{cz}_1+\mathrm{d}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}\right|=1\)
\(\left|\frac{1(1)+(-2)(2)+2(3)+\lambda}{\sqrt{1+4+4}}\right|=1\)
\(\left|\frac{1-4+6+\lambda}{3}\right|=1\)
\(3+\lambda= \pm 3\)
\(\therefore 3+\lambda=3 \quad\) or \(\quad 3+\lambda=-3\)
\(\lambda=0 \quad\) or \(\quad \lambda=-6\)
Hence, the equation of planes are \(x-2 y+2 z=0\) and \(x-2 y+2 z-6=0\)
MHT CET-2019
Three Dimensional Geometry
121258
If the foot of the perpendicular drawn form the point \((0,0,0)\) to the plane is \((4,-2,-5)\), then the equation of the plane is
1 \(4 x-2 y+5 z=-5\)
2 \(4 x-2 y-5 z=45\)
3 \(4 x+2 y-5 z=37\)
4 \(4 x+2 y+5 z=-13\)
Explanation:
B The required plane passes through the point \(\mathrm{M}(4,-2,-5)\) and is perpendicular to \(\overline{\mathrm{OM}}\) \(\therefore \overline{\mathrm{a}}=4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}\) and \(\overline{\mathrm{OM}}=\overline{\mathrm{n}}=4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}\), which is normal to the plane. \(\therefore \overline{\mathrm{a}}=\overline{\mathrm{n}}\) \(\therefore\) The equation of the plane is \(\overline{\mathrm{r}} \cdot \overline{\mathrm{n}}=\overline{\mathrm{a}} \cdot \overline{\mathrm{n}}\) \(\therefore \overline{\mathrm{r}} \cdot(4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \mathrm{k})=(4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}) \cdot(4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})\) \(\quad=16+4+25\) \(\overline{\mathrm{r}} \cdot(4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})=45\) \(\text { Taking } \overline{\mathrm{r}}=x \hat{\mathrm{i}}+\mathrm{y}+\mathrm{j} \hat{\mathrm{k}}\) \((x \hat{\mathrm{i}}+\mathrm{y}+\mathrm{j} \hat{\mathrm{k}}) \cdot(4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})=45 \Rightarrow 4 \mathrm{x}-2 \mathrm{y}-5 \mathrm{z}=45\)Taking \(\overline{\mathrm{r}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}\)
MHT CET-2019
Three Dimensional Geometry
121260
If lines \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\) and \(x-3=\frac{y-k}{2}=z\) interest, then the value of \(k\) is
1 \(\frac{9}{2}\)
2 \(\frac{1}{2}\)
3 \(\frac{5}{2}\)
4 \(\frac{7}{2}\)
Explanation:
A Points on the line are \((1,-1,1)\) and \((3, \mathrm{k}, 0)\) and direction ratios or line are 2,3,4 and 1,2,1.
Since lines intersect, then lies are coplanar,
\(\therefore\left|\begin{array}{ccc}\mathrm{x}_2-\mathrm{x}_1 & \mathrm{y}_2-\mathrm{y}_1 & \mathrm{z}_2-\mathrm{z}_1 \\ \mathrm{a}_1 & \mathrm{~b}_1 & \mathrm{c}_1 \\ \mathrm{a}_2 & \mathrm{~b}_2 & \mathrm{c}_2\end{array}\right|=0 \Rightarrow\left|\begin{array}{ccc}2 & \mathrm{k}+1 & -1 \\ 2 & 3 & 4 \\ 1 & 2 & 1\end{array}\right|=0\)
\(\therefore 2(-5)-(\mathrm{k}+1)(-2)-1(1)=0 \Rightarrow-11+2 \mathrm{k}+2=0 \Rightarrow \mathrm{k}=\frac{9}{2}\)
MHT CET-2018
Three Dimensional Geometry
121261
If planes \(x-c y-b y=0, c x-y+a z=0\) and \(b x\) \(+a y-z=0\) pass through a straight line, then \(\mathbf{a}^2+\mathbf{b}^2+\mathbf{c}^2=\)
1 \(1-\mathrm{abc}\)
2 abc -1
3 1-2abc
4 \(2 \mathrm{abc}-1\)
Explanation:
C Given, Planes are -
\(x-c y-b z=0\)
\(c x-y+a z=0\)
\(b x+a y-z=0\)
\(\therefore\) Planes are concurrent
\(\left|\begin{array}{ccc}1 & -c & -b \\ c & -1 & a \\ b & a & -1\end{array}\right|=0\)
\(1\left(1-a^2\right)+c(-c-a b)-b(a c+b)=0\)
\(1-a^2-c^2-a b c-a b c-b^2=0 \Rightarrow a^2+b^2+c^2=1-2 a b c\)
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Three Dimensional Geometry
121256
If the line passing through \(A(\lambda, 2,3)\) and \(B(1,2, \mu)\) is parallel to the line \(\mathbf{x}-\mathbf{1}=\mathbf{y}-\mathbf{2}=\mathbf{z}-\mathbf{3}\), then \(\lambda+\mu=\)
1 1
2 4
3 2
4 3
Explanation:
B The equation of line passing through \(\mathrm{A}(\lambda, 2,3)\) and \(\mathrm{B}(1,2, \mu)\) is \(\frac{x-\lambda}{\lambda-1}=\frac{y-2}{2-2}=\frac{z-3}{3-\mu}\) i.e. \(\frac{x-\lambda}{\lambda-1}=\frac{y-2}{0}=\frac{z-3}{3-\mu}\)... The line (i) is parallel to the line \(\mathrm{x}-1=\mathrm{y}-2=\mathrm{z}-3\) \(\therefore \lambda-1=0=3-\mu \Rightarrow \lambda=1\) and \(\mu=3\) Hence, \(\lambda+\mu=1+3=4\)
MHT CET-2019
Three Dimensional Geometry
121257
Equation of planes parallel to the plane \(x-2 y+2 z+4=0\) which are at a distance of one unit form the point \((1,2,3)\) are
1 \(x+2 y+2 z=6, x+2 y+2 z=0\)
2 \(x-2 y+2 z=0, x-2 y+2 z-6=0\)
3 \(x-2 y-6=0, x-2 y+z=6\)
4 \(x+2 y+2 z=-6, x+2 y+2 z=5\)
Explanation:
B The equation of planes parallel to the plane \(\mathrm{x}-2 \mathrm{y}+2 \mathrm{z}+4=0\) is \(\mathrm{x}-2 \mathrm{y}+2 \mathrm{z}+\lambda=0\)
The required planes are at a distance of one unit from the point \((1,2,3)\).
\(\left|\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{cz}_1+\mathrm{d}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}\right|=1\)
\(\left|\frac{1(1)+(-2)(2)+2(3)+\lambda}{\sqrt{1+4+4}}\right|=1\)
\(\left|\frac{1-4+6+\lambda}{3}\right|=1\)
\(3+\lambda= \pm 3\)
\(\therefore 3+\lambda=3 \quad\) or \(\quad 3+\lambda=-3\)
\(\lambda=0 \quad\) or \(\quad \lambda=-6\)
Hence, the equation of planes are \(x-2 y+2 z=0\) and \(x-2 y+2 z-6=0\)
MHT CET-2019
Three Dimensional Geometry
121258
If the foot of the perpendicular drawn form the point \((0,0,0)\) to the plane is \((4,-2,-5)\), then the equation of the plane is
1 \(4 x-2 y+5 z=-5\)
2 \(4 x-2 y-5 z=45\)
3 \(4 x+2 y-5 z=37\)
4 \(4 x+2 y+5 z=-13\)
Explanation:
B The required plane passes through the point \(\mathrm{M}(4,-2,-5)\) and is perpendicular to \(\overline{\mathrm{OM}}\) \(\therefore \overline{\mathrm{a}}=4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}\) and \(\overline{\mathrm{OM}}=\overline{\mathrm{n}}=4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}\), which is normal to the plane. \(\therefore \overline{\mathrm{a}}=\overline{\mathrm{n}}\) \(\therefore\) The equation of the plane is \(\overline{\mathrm{r}} \cdot \overline{\mathrm{n}}=\overline{\mathrm{a}} \cdot \overline{\mathrm{n}}\) \(\therefore \overline{\mathrm{r}} \cdot(4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \mathrm{k})=(4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}) \cdot(4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})\) \(\quad=16+4+25\) \(\overline{\mathrm{r}} \cdot(4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})=45\) \(\text { Taking } \overline{\mathrm{r}}=x \hat{\mathrm{i}}+\mathrm{y}+\mathrm{j} \hat{\mathrm{k}}\) \((x \hat{\mathrm{i}}+\mathrm{y}+\mathrm{j} \hat{\mathrm{k}}) \cdot(4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})=45 \Rightarrow 4 \mathrm{x}-2 \mathrm{y}-5 \mathrm{z}=45\)Taking \(\overline{\mathrm{r}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}\)
MHT CET-2019
Three Dimensional Geometry
121260
If lines \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\) and \(x-3=\frac{y-k}{2}=z\) interest, then the value of \(k\) is
1 \(\frac{9}{2}\)
2 \(\frac{1}{2}\)
3 \(\frac{5}{2}\)
4 \(\frac{7}{2}\)
Explanation:
A Points on the line are \((1,-1,1)\) and \((3, \mathrm{k}, 0)\) and direction ratios or line are 2,3,4 and 1,2,1.
Since lines intersect, then lies are coplanar,
\(\therefore\left|\begin{array}{ccc}\mathrm{x}_2-\mathrm{x}_1 & \mathrm{y}_2-\mathrm{y}_1 & \mathrm{z}_2-\mathrm{z}_1 \\ \mathrm{a}_1 & \mathrm{~b}_1 & \mathrm{c}_1 \\ \mathrm{a}_2 & \mathrm{~b}_2 & \mathrm{c}_2\end{array}\right|=0 \Rightarrow\left|\begin{array}{ccc}2 & \mathrm{k}+1 & -1 \\ 2 & 3 & 4 \\ 1 & 2 & 1\end{array}\right|=0\)
\(\therefore 2(-5)-(\mathrm{k}+1)(-2)-1(1)=0 \Rightarrow-11+2 \mathrm{k}+2=0 \Rightarrow \mathrm{k}=\frac{9}{2}\)
MHT CET-2018
Three Dimensional Geometry
121261
If planes \(x-c y-b y=0, c x-y+a z=0\) and \(b x\) \(+a y-z=0\) pass through a straight line, then \(\mathbf{a}^2+\mathbf{b}^2+\mathbf{c}^2=\)
1 \(1-\mathrm{abc}\)
2 abc -1
3 1-2abc
4 \(2 \mathrm{abc}-1\)
Explanation:
C Given, Planes are -
\(x-c y-b z=0\)
\(c x-y+a z=0\)
\(b x+a y-z=0\)
\(\therefore\) Planes are concurrent
\(\left|\begin{array}{ccc}1 & -c & -b \\ c & -1 & a \\ b & a & -1\end{array}\right|=0\)
\(1\left(1-a^2\right)+c(-c-a b)-b(a c+b)=0\)
\(1-a^2-c^2-a b c-a b c-b^2=0 \Rightarrow a^2+b^2+c^2=1-2 a b c\)
121256
If the line passing through \(A(\lambda, 2,3)\) and \(B(1,2, \mu)\) is parallel to the line \(\mathbf{x}-\mathbf{1}=\mathbf{y}-\mathbf{2}=\mathbf{z}-\mathbf{3}\), then \(\lambda+\mu=\)
1 1
2 4
3 2
4 3
Explanation:
B The equation of line passing through \(\mathrm{A}(\lambda, 2,3)\) and \(\mathrm{B}(1,2, \mu)\) is \(\frac{x-\lambda}{\lambda-1}=\frac{y-2}{2-2}=\frac{z-3}{3-\mu}\) i.e. \(\frac{x-\lambda}{\lambda-1}=\frac{y-2}{0}=\frac{z-3}{3-\mu}\)... The line (i) is parallel to the line \(\mathrm{x}-1=\mathrm{y}-2=\mathrm{z}-3\) \(\therefore \lambda-1=0=3-\mu \Rightarrow \lambda=1\) and \(\mu=3\) Hence, \(\lambda+\mu=1+3=4\)
MHT CET-2019
Three Dimensional Geometry
121257
Equation of planes parallel to the plane \(x-2 y+2 z+4=0\) which are at a distance of one unit form the point \((1,2,3)\) are
1 \(x+2 y+2 z=6, x+2 y+2 z=0\)
2 \(x-2 y+2 z=0, x-2 y+2 z-6=0\)
3 \(x-2 y-6=0, x-2 y+z=6\)
4 \(x+2 y+2 z=-6, x+2 y+2 z=5\)
Explanation:
B The equation of planes parallel to the plane \(\mathrm{x}-2 \mathrm{y}+2 \mathrm{z}+4=0\) is \(\mathrm{x}-2 \mathrm{y}+2 \mathrm{z}+\lambda=0\)
The required planes are at a distance of one unit from the point \((1,2,3)\).
\(\left|\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{cz}_1+\mathrm{d}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}\right|=1\)
\(\left|\frac{1(1)+(-2)(2)+2(3)+\lambda}{\sqrt{1+4+4}}\right|=1\)
\(\left|\frac{1-4+6+\lambda}{3}\right|=1\)
\(3+\lambda= \pm 3\)
\(\therefore 3+\lambda=3 \quad\) or \(\quad 3+\lambda=-3\)
\(\lambda=0 \quad\) or \(\quad \lambda=-6\)
Hence, the equation of planes are \(x-2 y+2 z=0\) and \(x-2 y+2 z-6=0\)
MHT CET-2019
Three Dimensional Geometry
121258
If the foot of the perpendicular drawn form the point \((0,0,0)\) to the plane is \((4,-2,-5)\), then the equation of the plane is
1 \(4 x-2 y+5 z=-5\)
2 \(4 x-2 y-5 z=45\)
3 \(4 x+2 y-5 z=37\)
4 \(4 x+2 y+5 z=-13\)
Explanation:
B The required plane passes through the point \(\mathrm{M}(4,-2,-5)\) and is perpendicular to \(\overline{\mathrm{OM}}\) \(\therefore \overline{\mathrm{a}}=4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}\) and \(\overline{\mathrm{OM}}=\overline{\mathrm{n}}=4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}\), which is normal to the plane. \(\therefore \overline{\mathrm{a}}=\overline{\mathrm{n}}\) \(\therefore\) The equation of the plane is \(\overline{\mathrm{r}} \cdot \overline{\mathrm{n}}=\overline{\mathrm{a}} \cdot \overline{\mathrm{n}}\) \(\therefore \overline{\mathrm{r}} \cdot(4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \mathrm{k})=(4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}) \cdot(4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})\) \(\quad=16+4+25\) \(\overline{\mathrm{r}} \cdot(4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})=45\) \(\text { Taking } \overline{\mathrm{r}}=x \hat{\mathrm{i}}+\mathrm{y}+\mathrm{j} \hat{\mathrm{k}}\) \((x \hat{\mathrm{i}}+\mathrm{y}+\mathrm{j} \hat{\mathrm{k}}) \cdot(4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})=45 \Rightarrow 4 \mathrm{x}-2 \mathrm{y}-5 \mathrm{z}=45\)Taking \(\overline{\mathrm{r}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}\)
MHT CET-2019
Three Dimensional Geometry
121260
If lines \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\) and \(x-3=\frac{y-k}{2}=z\) interest, then the value of \(k\) is
1 \(\frac{9}{2}\)
2 \(\frac{1}{2}\)
3 \(\frac{5}{2}\)
4 \(\frac{7}{2}\)
Explanation:
A Points on the line are \((1,-1,1)\) and \((3, \mathrm{k}, 0)\) and direction ratios or line are 2,3,4 and 1,2,1.
Since lines intersect, then lies are coplanar,
\(\therefore\left|\begin{array}{ccc}\mathrm{x}_2-\mathrm{x}_1 & \mathrm{y}_2-\mathrm{y}_1 & \mathrm{z}_2-\mathrm{z}_1 \\ \mathrm{a}_1 & \mathrm{~b}_1 & \mathrm{c}_1 \\ \mathrm{a}_2 & \mathrm{~b}_2 & \mathrm{c}_2\end{array}\right|=0 \Rightarrow\left|\begin{array}{ccc}2 & \mathrm{k}+1 & -1 \\ 2 & 3 & 4 \\ 1 & 2 & 1\end{array}\right|=0\)
\(\therefore 2(-5)-(\mathrm{k}+1)(-2)-1(1)=0 \Rightarrow-11+2 \mathrm{k}+2=0 \Rightarrow \mathrm{k}=\frac{9}{2}\)
MHT CET-2018
Three Dimensional Geometry
121261
If planes \(x-c y-b y=0, c x-y+a z=0\) and \(b x\) \(+a y-z=0\) pass through a straight line, then \(\mathbf{a}^2+\mathbf{b}^2+\mathbf{c}^2=\)
1 \(1-\mathrm{abc}\)
2 abc -1
3 1-2abc
4 \(2 \mathrm{abc}-1\)
Explanation:
C Given, Planes are -
\(x-c y-b z=0\)
\(c x-y+a z=0\)
\(b x+a y-z=0\)
\(\therefore\) Planes are concurrent
\(\left|\begin{array}{ccc}1 & -c & -b \\ c & -1 & a \\ b & a & -1\end{array}\right|=0\)
\(1\left(1-a^2\right)+c(-c-a b)-b(a c+b)=0\)
\(1-a^2-c^2-a b c-a b c-b^2=0 \Rightarrow a^2+b^2+c^2=1-2 a b c\)
121256
If the line passing through \(A(\lambda, 2,3)\) and \(B(1,2, \mu)\) is parallel to the line \(\mathbf{x}-\mathbf{1}=\mathbf{y}-\mathbf{2}=\mathbf{z}-\mathbf{3}\), then \(\lambda+\mu=\)
1 1
2 4
3 2
4 3
Explanation:
B The equation of line passing through \(\mathrm{A}(\lambda, 2,3)\) and \(\mathrm{B}(1,2, \mu)\) is \(\frac{x-\lambda}{\lambda-1}=\frac{y-2}{2-2}=\frac{z-3}{3-\mu}\) i.e. \(\frac{x-\lambda}{\lambda-1}=\frac{y-2}{0}=\frac{z-3}{3-\mu}\)... The line (i) is parallel to the line \(\mathrm{x}-1=\mathrm{y}-2=\mathrm{z}-3\) \(\therefore \lambda-1=0=3-\mu \Rightarrow \lambda=1\) and \(\mu=3\) Hence, \(\lambda+\mu=1+3=4\)
MHT CET-2019
Three Dimensional Geometry
121257
Equation of planes parallel to the plane \(x-2 y+2 z+4=0\) which are at a distance of one unit form the point \((1,2,3)\) are
1 \(x+2 y+2 z=6, x+2 y+2 z=0\)
2 \(x-2 y+2 z=0, x-2 y+2 z-6=0\)
3 \(x-2 y-6=0, x-2 y+z=6\)
4 \(x+2 y+2 z=-6, x+2 y+2 z=5\)
Explanation:
B The equation of planes parallel to the plane \(\mathrm{x}-2 \mathrm{y}+2 \mathrm{z}+4=0\) is \(\mathrm{x}-2 \mathrm{y}+2 \mathrm{z}+\lambda=0\)
The required planes are at a distance of one unit from the point \((1,2,3)\).
\(\left|\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{cz}_1+\mathrm{d}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}\right|=1\)
\(\left|\frac{1(1)+(-2)(2)+2(3)+\lambda}{\sqrt{1+4+4}}\right|=1\)
\(\left|\frac{1-4+6+\lambda}{3}\right|=1\)
\(3+\lambda= \pm 3\)
\(\therefore 3+\lambda=3 \quad\) or \(\quad 3+\lambda=-3\)
\(\lambda=0 \quad\) or \(\quad \lambda=-6\)
Hence, the equation of planes are \(x-2 y+2 z=0\) and \(x-2 y+2 z-6=0\)
MHT CET-2019
Three Dimensional Geometry
121258
If the foot of the perpendicular drawn form the point \((0,0,0)\) to the plane is \((4,-2,-5)\), then the equation of the plane is
1 \(4 x-2 y+5 z=-5\)
2 \(4 x-2 y-5 z=45\)
3 \(4 x+2 y-5 z=37\)
4 \(4 x+2 y+5 z=-13\)
Explanation:
B The required plane passes through the point \(\mathrm{M}(4,-2,-5)\) and is perpendicular to \(\overline{\mathrm{OM}}\) \(\therefore \overline{\mathrm{a}}=4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}\) and \(\overline{\mathrm{OM}}=\overline{\mathrm{n}}=4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}\), which is normal to the plane. \(\therefore \overline{\mathrm{a}}=\overline{\mathrm{n}}\) \(\therefore\) The equation of the plane is \(\overline{\mathrm{r}} \cdot \overline{\mathrm{n}}=\overline{\mathrm{a}} \cdot \overline{\mathrm{n}}\) \(\therefore \overline{\mathrm{r}} \cdot(4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \mathrm{k})=(4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}) \cdot(4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})\) \(\quad=16+4+25\) \(\overline{\mathrm{r}} \cdot(4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})=45\) \(\text { Taking } \overline{\mathrm{r}}=x \hat{\mathrm{i}}+\mathrm{y}+\mathrm{j} \hat{\mathrm{k}}\) \((x \hat{\mathrm{i}}+\mathrm{y}+\mathrm{j} \hat{\mathrm{k}}) \cdot(4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})=45 \Rightarrow 4 \mathrm{x}-2 \mathrm{y}-5 \mathrm{z}=45\)Taking \(\overline{\mathrm{r}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}\)
MHT CET-2019
Three Dimensional Geometry
121260
If lines \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\) and \(x-3=\frac{y-k}{2}=z\) interest, then the value of \(k\) is
1 \(\frac{9}{2}\)
2 \(\frac{1}{2}\)
3 \(\frac{5}{2}\)
4 \(\frac{7}{2}\)
Explanation:
A Points on the line are \((1,-1,1)\) and \((3, \mathrm{k}, 0)\) and direction ratios or line are 2,3,4 and 1,2,1.
Since lines intersect, then lies are coplanar,
\(\therefore\left|\begin{array}{ccc}\mathrm{x}_2-\mathrm{x}_1 & \mathrm{y}_2-\mathrm{y}_1 & \mathrm{z}_2-\mathrm{z}_1 \\ \mathrm{a}_1 & \mathrm{~b}_1 & \mathrm{c}_1 \\ \mathrm{a}_2 & \mathrm{~b}_2 & \mathrm{c}_2\end{array}\right|=0 \Rightarrow\left|\begin{array}{ccc}2 & \mathrm{k}+1 & -1 \\ 2 & 3 & 4 \\ 1 & 2 & 1\end{array}\right|=0\)
\(\therefore 2(-5)-(\mathrm{k}+1)(-2)-1(1)=0 \Rightarrow-11+2 \mathrm{k}+2=0 \Rightarrow \mathrm{k}=\frac{9}{2}\)
MHT CET-2018
Three Dimensional Geometry
121261
If planes \(x-c y-b y=0, c x-y+a z=0\) and \(b x\) \(+a y-z=0\) pass through a straight line, then \(\mathbf{a}^2+\mathbf{b}^2+\mathbf{c}^2=\)
1 \(1-\mathrm{abc}\)
2 abc -1
3 1-2abc
4 \(2 \mathrm{abc}-1\)
Explanation:
C Given, Planes are -
\(x-c y-b z=0\)
\(c x-y+a z=0\)
\(b x+a y-z=0\)
\(\therefore\) Planes are concurrent
\(\left|\begin{array}{ccc}1 & -c & -b \\ c & -1 & a \\ b & a & -1\end{array}\right|=0\)
\(1\left(1-a^2\right)+c(-c-a b)-b(a c+b)=0\)
\(1-a^2-c^2-a b c-a b c-b^2=0 \Rightarrow a^2+b^2+c^2=1-2 a b c\)
