121126
If the acute angle between the line with direction ratios \(1,-1\), a and \(2,1,-1\) is \(60^{\circ}\), then the value of ' \(a\) ' is
1 1
2 -2
3 2
4 -1
Explanation:
B Given, \(a_1=1, b_1=-1, c_1=a\) and \(a_2=2, b_2=1, c_2=-1\) and \(\theta=\) \(60^{\circ}\) We know that,
\(\cos \theta=\left \vert\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right \vert \Rightarrow \Rightarrow\)
\(\Rightarrow \cos 60^{\circ}=\left \vert\frac{2+(-1)+(-a)}{\sqrt{1+1+a^2} \sqrt{4+1+1}}\right \vert\)
\(\Rightarrow \quad \frac{1}{2}=\left \vert\frac{1-a}{\sqrt{2+a^2} \sqrt{6}}\right \vert\)
On squaring both sides, we get -
\(\frac{1}{4}=\frac{(1-a)^2}{6\left(2+a^2\right)} \Rightarrow \frac{1}{4}=\frac{1-2 a+a^2}{12+6 a^2}\)
\(\Rightarrow \quad 12+6 a^2=4-8 a+4 a^2\)
\(\Rightarrow \quad 2 a^2+8 a+8=0\)
\(a^2+4 a+4=0\)
\((a+2)^2=0 \Rightarrow a=-2\)
MHT CET-2019
Three Dimensional Geometry
121134
The direction ratios of the line which is perpendicular to the lines
\(\frac{x-7}{2}=\frac{y+17}{-3}=\frac{z-6}{1}\) and \(\frac{x+5}{1}=\frac{y+3}{2}=\frac{z-4}{-2}\) are
1 \(\langle 4,5,7\rangle\)
2 \(\langle 4,-5,7\rangle\)
3 \(\langle 4,-5,-7\rangle\)
4 \(\langle-4,5,7\rangle\)
Explanation:
A Given- \(\frac{x-7}{2}=\frac{y+17}{-3}=\frac{z-6}{1}\)
and \(\quad \frac{x+5}{1}=\frac{y+3}{2}=\frac{z-4}{-2}\)
Direction ratios of the given lines are proportional to 2, \(-3,1\) and \(1,2-2\), which is parallel to the vectors-
\(\vec{a}_1=2 \hat{i}-3 \hat{j}+\hat{k} \text { and } \vec{a}_2=\hat{i}+2 \hat{j}-2 \hat{k}\)
\(\therefore\) These line are perpendicular, Then,
\(\overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{a}}_1 \times \overrightarrow{\mathrm{a}}_2\)
\(=\left \vert\begin{array}{rrr}\hat{i} & \hat{j} & \hat{\mathrm{k}} \\ 2 & -3 & 1 \\ 1 & 2 & -2\end{array}\right \vert\)
\(=4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}\)Therefore, direction ratios of the line perpendicular to given two line are proportional to \((4,5,7)\)
COMEDK-2015
Three Dimensional Geometry
121146
If the line of intersection of the planes \(a x+b y=\) 3 and \(a x+b y+c z=0, a>0\) makes and angle \(30^{\circ}\) with the plane \(\mathrm{y}-\mathrm{z}+\mathbf{2}=\mathbf{0}\), then the direction cosines of the line are :
1 \(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\)
2 \(\frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}}, 0\)
3 \(\frac{1}{\sqrt{5}},-\frac{2}{\sqrt{5}}, 0\)
4 \(\frac{1}{2},-\frac{\sqrt{3}}{2}, 0\)
Explanation:
B Planes are \(a x+b y-3=0\) and \(a x+b y+c z=0\) Line of intersection is parallel to \(\mathrm{n}_1\) and \(\mathrm{n}_2\) Now,
Now, \(\quad \mathrm{n}_1 \times \mathrm{n}_2=\left \vert\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ \mathrm{a} & \mathrm{b} & 0 \\ \mathrm{a} & \mathrm{b} & \mathrm{c}\end{array}\right \vert\)
\(=(\mathrm{bc}) \hat{\mathrm{i}}-(\mathrm{ac}) \hat{\mathrm{j}}\)
\(\therefore\) Direction cosines of line of intersection can be-
\(\left(\frac{\mathrm{bc}}{\sqrt{\mathrm{b}^2 \mathrm{c}^2+\mathrm{a}^2 \mathrm{c}^2}}, \frac{-\mathrm{ac}}{\sqrt{\mathrm{b}^2 \mathrm{c}^2+\mathrm{a}^2 \mathrm{c}^2}}, 0\right)\)
This line makes \(30^{\circ}\) with plane \(y-z+2=0\)
\(\therefore\) It makes \(60^{\circ}\) with normal \((0,1,-1)\)
\(\therefore \quad \cos 60^{\circ}=\frac{-\mathrm{ac}}{\sqrt{2} \sqrt{\mathrm{b}^2 \mathrm{c}^2+\mathrm{a}^2 \mathrm{c}^2}}\)
Squaring on both sides-
\(\frac{1}{4} =\frac{a^2 c^2}{(2)\left(b^2 c^2+a^2 c^2\right)}\)
\(a^2 =b^2\)
Direction cosines of line of intersection are-
\(\left(\frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}}, 0\right)\)
121126
If the acute angle between the line with direction ratios \(1,-1\), a and \(2,1,-1\) is \(60^{\circ}\), then the value of ' \(a\) ' is
1 1
2 -2
3 2
4 -1
Explanation:
B Given, \(a_1=1, b_1=-1, c_1=a\) and \(a_2=2, b_2=1, c_2=-1\) and \(\theta=\) \(60^{\circ}\) We know that,
\(\cos \theta=\left \vert\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right \vert \Rightarrow \Rightarrow\)
\(\Rightarrow \cos 60^{\circ}=\left \vert\frac{2+(-1)+(-a)}{\sqrt{1+1+a^2} \sqrt{4+1+1}}\right \vert\)
\(\Rightarrow \quad \frac{1}{2}=\left \vert\frac{1-a}{\sqrt{2+a^2} \sqrt{6}}\right \vert\)
On squaring both sides, we get -
\(\frac{1}{4}=\frac{(1-a)^2}{6\left(2+a^2\right)} \Rightarrow \frac{1}{4}=\frac{1-2 a+a^2}{12+6 a^2}\)
\(\Rightarrow \quad 12+6 a^2=4-8 a+4 a^2\)
\(\Rightarrow \quad 2 a^2+8 a+8=0\)
\(a^2+4 a+4=0\)
\((a+2)^2=0 \Rightarrow a=-2\)
MHT CET-2019
Three Dimensional Geometry
121134
The direction ratios of the line which is perpendicular to the lines
\(\frac{x-7}{2}=\frac{y+17}{-3}=\frac{z-6}{1}\) and \(\frac{x+5}{1}=\frac{y+3}{2}=\frac{z-4}{-2}\) are
1 \(\langle 4,5,7\rangle\)
2 \(\langle 4,-5,7\rangle\)
3 \(\langle 4,-5,-7\rangle\)
4 \(\langle-4,5,7\rangle\)
Explanation:
A Given- \(\frac{x-7}{2}=\frac{y+17}{-3}=\frac{z-6}{1}\)
and \(\quad \frac{x+5}{1}=\frac{y+3}{2}=\frac{z-4}{-2}\)
Direction ratios of the given lines are proportional to 2, \(-3,1\) and \(1,2-2\), which is parallel to the vectors-
\(\vec{a}_1=2 \hat{i}-3 \hat{j}+\hat{k} \text { and } \vec{a}_2=\hat{i}+2 \hat{j}-2 \hat{k}\)
\(\therefore\) These line are perpendicular, Then,
\(\overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{a}}_1 \times \overrightarrow{\mathrm{a}}_2\)
\(=\left \vert\begin{array}{rrr}\hat{i} & \hat{j} & \hat{\mathrm{k}} \\ 2 & -3 & 1 \\ 1 & 2 & -2\end{array}\right \vert\)
\(=4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}\)Therefore, direction ratios of the line perpendicular to given two line are proportional to \((4,5,7)\)
COMEDK-2015
Three Dimensional Geometry
121146
If the line of intersection of the planes \(a x+b y=\) 3 and \(a x+b y+c z=0, a>0\) makes and angle \(30^{\circ}\) with the plane \(\mathrm{y}-\mathrm{z}+\mathbf{2}=\mathbf{0}\), then the direction cosines of the line are :
1 \(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\)
2 \(\frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}}, 0\)
3 \(\frac{1}{\sqrt{5}},-\frac{2}{\sqrt{5}}, 0\)
4 \(\frac{1}{2},-\frac{\sqrt{3}}{2}, 0\)
Explanation:
B Planes are \(a x+b y-3=0\) and \(a x+b y+c z=0\) Line of intersection is parallel to \(\mathrm{n}_1\) and \(\mathrm{n}_2\) Now,
Now, \(\quad \mathrm{n}_1 \times \mathrm{n}_2=\left \vert\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ \mathrm{a} & \mathrm{b} & 0 \\ \mathrm{a} & \mathrm{b} & \mathrm{c}\end{array}\right \vert\)
\(=(\mathrm{bc}) \hat{\mathrm{i}}-(\mathrm{ac}) \hat{\mathrm{j}}\)
\(\therefore\) Direction cosines of line of intersection can be-
\(\left(\frac{\mathrm{bc}}{\sqrt{\mathrm{b}^2 \mathrm{c}^2+\mathrm{a}^2 \mathrm{c}^2}}, \frac{-\mathrm{ac}}{\sqrt{\mathrm{b}^2 \mathrm{c}^2+\mathrm{a}^2 \mathrm{c}^2}}, 0\right)\)
This line makes \(30^{\circ}\) with plane \(y-z+2=0\)
\(\therefore\) It makes \(60^{\circ}\) with normal \((0,1,-1)\)
\(\therefore \quad \cos 60^{\circ}=\frac{-\mathrm{ac}}{\sqrt{2} \sqrt{\mathrm{b}^2 \mathrm{c}^2+\mathrm{a}^2 \mathrm{c}^2}}\)
Squaring on both sides-
\(\frac{1}{4} =\frac{a^2 c^2}{(2)\left(b^2 c^2+a^2 c^2\right)}\)
\(a^2 =b^2\)
Direction cosines of line of intersection are-
\(\left(\frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}}, 0\right)\)
121126
If the acute angle between the line with direction ratios \(1,-1\), a and \(2,1,-1\) is \(60^{\circ}\), then the value of ' \(a\) ' is
1 1
2 -2
3 2
4 -1
Explanation:
B Given, \(a_1=1, b_1=-1, c_1=a\) and \(a_2=2, b_2=1, c_2=-1\) and \(\theta=\) \(60^{\circ}\) We know that,
\(\cos \theta=\left \vert\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right \vert \Rightarrow \Rightarrow\)
\(\Rightarrow \cos 60^{\circ}=\left \vert\frac{2+(-1)+(-a)}{\sqrt{1+1+a^2} \sqrt{4+1+1}}\right \vert\)
\(\Rightarrow \quad \frac{1}{2}=\left \vert\frac{1-a}{\sqrt{2+a^2} \sqrt{6}}\right \vert\)
On squaring both sides, we get -
\(\frac{1}{4}=\frac{(1-a)^2}{6\left(2+a^2\right)} \Rightarrow \frac{1}{4}=\frac{1-2 a+a^2}{12+6 a^2}\)
\(\Rightarrow \quad 12+6 a^2=4-8 a+4 a^2\)
\(\Rightarrow \quad 2 a^2+8 a+8=0\)
\(a^2+4 a+4=0\)
\((a+2)^2=0 \Rightarrow a=-2\)
MHT CET-2019
Three Dimensional Geometry
121134
The direction ratios of the line which is perpendicular to the lines
\(\frac{x-7}{2}=\frac{y+17}{-3}=\frac{z-6}{1}\) and \(\frac{x+5}{1}=\frac{y+3}{2}=\frac{z-4}{-2}\) are
1 \(\langle 4,5,7\rangle\)
2 \(\langle 4,-5,7\rangle\)
3 \(\langle 4,-5,-7\rangle\)
4 \(\langle-4,5,7\rangle\)
Explanation:
A Given- \(\frac{x-7}{2}=\frac{y+17}{-3}=\frac{z-6}{1}\)
and \(\quad \frac{x+5}{1}=\frac{y+3}{2}=\frac{z-4}{-2}\)
Direction ratios of the given lines are proportional to 2, \(-3,1\) and \(1,2-2\), which is parallel to the vectors-
\(\vec{a}_1=2 \hat{i}-3 \hat{j}+\hat{k} \text { and } \vec{a}_2=\hat{i}+2 \hat{j}-2 \hat{k}\)
\(\therefore\) These line are perpendicular, Then,
\(\overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{a}}_1 \times \overrightarrow{\mathrm{a}}_2\)
\(=\left \vert\begin{array}{rrr}\hat{i} & \hat{j} & \hat{\mathrm{k}} \\ 2 & -3 & 1 \\ 1 & 2 & -2\end{array}\right \vert\)
\(=4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}\)Therefore, direction ratios of the line perpendicular to given two line are proportional to \((4,5,7)\)
COMEDK-2015
Three Dimensional Geometry
121146
If the line of intersection of the planes \(a x+b y=\) 3 and \(a x+b y+c z=0, a>0\) makes and angle \(30^{\circ}\) with the plane \(\mathrm{y}-\mathrm{z}+\mathbf{2}=\mathbf{0}\), then the direction cosines of the line are :
1 \(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\)
2 \(\frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}}, 0\)
3 \(\frac{1}{\sqrt{5}},-\frac{2}{\sqrt{5}}, 0\)
4 \(\frac{1}{2},-\frac{\sqrt{3}}{2}, 0\)
Explanation:
B Planes are \(a x+b y-3=0\) and \(a x+b y+c z=0\) Line of intersection is parallel to \(\mathrm{n}_1\) and \(\mathrm{n}_2\) Now,
Now, \(\quad \mathrm{n}_1 \times \mathrm{n}_2=\left \vert\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ \mathrm{a} & \mathrm{b} & 0 \\ \mathrm{a} & \mathrm{b} & \mathrm{c}\end{array}\right \vert\)
\(=(\mathrm{bc}) \hat{\mathrm{i}}-(\mathrm{ac}) \hat{\mathrm{j}}\)
\(\therefore\) Direction cosines of line of intersection can be-
\(\left(\frac{\mathrm{bc}}{\sqrt{\mathrm{b}^2 \mathrm{c}^2+\mathrm{a}^2 \mathrm{c}^2}}, \frac{-\mathrm{ac}}{\sqrt{\mathrm{b}^2 \mathrm{c}^2+\mathrm{a}^2 \mathrm{c}^2}}, 0\right)\)
This line makes \(30^{\circ}\) with plane \(y-z+2=0\)
\(\therefore\) It makes \(60^{\circ}\) with normal \((0,1,-1)\)
\(\therefore \quad \cos 60^{\circ}=\frac{-\mathrm{ac}}{\sqrt{2} \sqrt{\mathrm{b}^2 \mathrm{c}^2+\mathrm{a}^2 \mathrm{c}^2}}\)
Squaring on both sides-
\(\frac{1}{4} =\frac{a^2 c^2}{(2)\left(b^2 c^2+a^2 c^2\right)}\)
\(a^2 =b^2\)
Direction cosines of line of intersection are-
\(\left(\frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}}, 0\right)\)
121126
If the acute angle between the line with direction ratios \(1,-1\), a and \(2,1,-1\) is \(60^{\circ}\), then the value of ' \(a\) ' is
1 1
2 -2
3 2
4 -1
Explanation:
B Given, \(a_1=1, b_1=-1, c_1=a\) and \(a_2=2, b_2=1, c_2=-1\) and \(\theta=\) \(60^{\circ}\) We know that,
\(\cos \theta=\left \vert\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right \vert \Rightarrow \Rightarrow\)
\(\Rightarrow \cos 60^{\circ}=\left \vert\frac{2+(-1)+(-a)}{\sqrt{1+1+a^2} \sqrt{4+1+1}}\right \vert\)
\(\Rightarrow \quad \frac{1}{2}=\left \vert\frac{1-a}{\sqrt{2+a^2} \sqrt{6}}\right \vert\)
On squaring both sides, we get -
\(\frac{1}{4}=\frac{(1-a)^2}{6\left(2+a^2\right)} \Rightarrow \frac{1}{4}=\frac{1-2 a+a^2}{12+6 a^2}\)
\(\Rightarrow \quad 12+6 a^2=4-8 a+4 a^2\)
\(\Rightarrow \quad 2 a^2+8 a+8=0\)
\(a^2+4 a+4=0\)
\((a+2)^2=0 \Rightarrow a=-2\)
MHT CET-2019
Three Dimensional Geometry
121134
The direction ratios of the line which is perpendicular to the lines
\(\frac{x-7}{2}=\frac{y+17}{-3}=\frac{z-6}{1}\) and \(\frac{x+5}{1}=\frac{y+3}{2}=\frac{z-4}{-2}\) are
1 \(\langle 4,5,7\rangle\)
2 \(\langle 4,-5,7\rangle\)
3 \(\langle 4,-5,-7\rangle\)
4 \(\langle-4,5,7\rangle\)
Explanation:
A Given- \(\frac{x-7}{2}=\frac{y+17}{-3}=\frac{z-6}{1}\)
and \(\quad \frac{x+5}{1}=\frac{y+3}{2}=\frac{z-4}{-2}\)
Direction ratios of the given lines are proportional to 2, \(-3,1\) and \(1,2-2\), which is parallel to the vectors-
\(\vec{a}_1=2 \hat{i}-3 \hat{j}+\hat{k} \text { and } \vec{a}_2=\hat{i}+2 \hat{j}-2 \hat{k}\)
\(\therefore\) These line are perpendicular, Then,
\(\overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{a}}_1 \times \overrightarrow{\mathrm{a}}_2\)
\(=\left \vert\begin{array}{rrr}\hat{i} & \hat{j} & \hat{\mathrm{k}} \\ 2 & -3 & 1 \\ 1 & 2 & -2\end{array}\right \vert\)
\(=4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}\)Therefore, direction ratios of the line perpendicular to given two line are proportional to \((4,5,7)\)
COMEDK-2015
Three Dimensional Geometry
121146
If the line of intersection of the planes \(a x+b y=\) 3 and \(a x+b y+c z=0, a>0\) makes and angle \(30^{\circ}\) with the plane \(\mathrm{y}-\mathrm{z}+\mathbf{2}=\mathbf{0}\), then the direction cosines of the line are :
1 \(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\)
2 \(\frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}}, 0\)
3 \(\frac{1}{\sqrt{5}},-\frac{2}{\sqrt{5}}, 0\)
4 \(\frac{1}{2},-\frac{\sqrt{3}}{2}, 0\)
Explanation:
B Planes are \(a x+b y-3=0\) and \(a x+b y+c z=0\) Line of intersection is parallel to \(\mathrm{n}_1\) and \(\mathrm{n}_2\) Now,
Now, \(\quad \mathrm{n}_1 \times \mathrm{n}_2=\left \vert\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ \mathrm{a} & \mathrm{b} & 0 \\ \mathrm{a} & \mathrm{b} & \mathrm{c}\end{array}\right \vert\)
\(=(\mathrm{bc}) \hat{\mathrm{i}}-(\mathrm{ac}) \hat{\mathrm{j}}\)
\(\therefore\) Direction cosines of line of intersection can be-
\(\left(\frac{\mathrm{bc}}{\sqrt{\mathrm{b}^2 \mathrm{c}^2+\mathrm{a}^2 \mathrm{c}^2}}, \frac{-\mathrm{ac}}{\sqrt{\mathrm{b}^2 \mathrm{c}^2+\mathrm{a}^2 \mathrm{c}^2}}, 0\right)\)
This line makes \(30^{\circ}\) with plane \(y-z+2=0\)
\(\therefore\) It makes \(60^{\circ}\) with normal \((0,1,-1)\)
\(\therefore \quad \cos 60^{\circ}=\frac{-\mathrm{ac}}{\sqrt{2} \sqrt{\mathrm{b}^2 \mathrm{c}^2+\mathrm{a}^2 \mathrm{c}^2}}\)
Squaring on both sides-
\(\frac{1}{4} =\frac{a^2 c^2}{(2)\left(b^2 c^2+a^2 c^2\right)}\)
\(a^2 =b^2\)
Direction cosines of line of intersection are-
\(\left(\frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}}, 0\right)\)
