121309
The angle between the lines \(\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}\) and \(\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-4}{1}\) is
1 \(\cos ^{-1}\left(\frac{1}{3}\right)\)
2 \(\cos ^{-1}\left(\frac{3}{4}\right)\)
3 \(\cos ^{-1}\left(\frac{1}{2}\right)\)
4 \(\cos ^{-1}\left(\frac{2}{3}\right)\)
Explanation:
D Given, Line, \(\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}\) and \(\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-4}{1}\)
\(\therefore \vec{a}=4 \hat{i}+\hat{j}+8 \hat{k} \text { and } \vec{b}=2 \hat{i}+2 \hat{j}+\hat{k}\)
\(\vec{a} \cdot \vec{b}=(4 \hat{i}+\hat{j}+8 \hat{k}) \cdot(2 \hat{i}+2 \hat{j}+\hat{k}) \quad(\because \hat{i} \cdot \hat{i}=\hat{j} \cdot \hat{j}=\hat{k} \cdot \hat{k}=1)\)
\(\vec{a} \cdot \vec{b}=(4 \times 2)+(1 \times 2)+(8 \times 1)=8+2+8=18\)
\(\quad \vert\vec{a} \vert=\sqrt{16+1+64}=\sqrt{81}=9\)
And, \(\quad \vert\vec{b} \vert=\sqrt{4+4+1}=\sqrt{9}=9\)
Let, \(\theta\) be the acute angle between the two given lines.
\(\cos \theta=\frac{\vec{a} \cdot \vec{b}}{ \vert\vec{a} \vert \vert\vec{b} \vert}=\frac{18}{9 \times 3}=\frac{2}{3}\)
\(\theta=\cos ^{-1}\left(\frac{2}{3}\right)\)
MHT CET-2020
Three Dimensional Geometry
121312
The angle between the two lines \(\frac{x-4}{1}=\frac{y+4}{2}=\frac{z+1}{2}\) and \(\frac{x+1}{2}=\frac{y+3}{2}=\frac{z-4}{-1}\) is
1 \(\cos ^{-1}\left(\frac{1}{3}\right)\)
2 \(\cos ^{-1}\left(\frac{4}{9}\right)\)
3 \(\cos ^{-1}\left(\frac{5}{9}\right)\)
4 \(\cos ^{-1}\left(\frac{2}{9}\right)\)
Explanation:
B Given, \(\mathrm{a}_1=2, \quad \mathrm{~b}_1=2 \quad \mathrm{c}_1=-1\)
\(\mathrm{a}_2=1, \quad \mathrm{~b}_2=2, \quad \mathrm{c}_2=2\)
We know that,
\(\cos \theta=\left \vert\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right \vert\)
Let, \(\theta\) be the required angle
\(\cos \theta=\left \vert\frac{1(2)+2(2)+2(-1)}{\sqrt{1+4+4} \sqrt{4+4+1}}\right \vert=\left \vert\frac{2+4-2}{3 \times 3}\right \vert\)
\(\cos \theta=\frac{4}{9}\)
\(\theta=\cos ^{-1}\left(\frac{4}{9}\right)\)
MHT CET-2020
Three Dimensional Geometry
121313
If the angle between the lines whose direction ratio are \(4,-3,5\) and \(3,4, \mathrm{k}\) is \(\frac{\pi}{3}\), then \(\mathrm{k}=\)
1 \(\pm 10\)
2 \(\pm 7\)
3 \(\pm 6\)
4 \(\pm 5\)
Explanation:
D The angle between two line, \(\cos \frac{\pi}{3}=\left \vert\frac{4(3)+(-3)(4)+5 \mathrm{k}}{\sqrt{4^2+(-3)^2+5^2} \sqrt{3^2+4^2+\mathrm{k}^2}}\right \vert\)
\(\therefore \frac{1}{2}=\left \vert\frac{5 \mathrm{k}}{5 \sqrt{2} \sqrt{25+\mathrm{k}^2}}\right \vert\)
On squaring both side we get-
\(\frac{1}{4}=\frac{25 \mathrm{k}^2}{50 \times\left(25+\mathrm{k}^2\right)}\)
\(\therefore 100 \mathrm{k}^2=50\left(25+\mathrm{k}^2\right) \Rightarrow \mathrm{k}^2=25 \Rightarrow \mathrm{k}= \pm 5\)
MHT CET-2020
Three Dimensional Geometry
121314
The angle between the line \(\frac{x-1}{2}=\frac{y+3}{1}=\frac{z+7}{2}\) and the plane \(\overline{\mathbf{r}} \cdot(6 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})=5\) is
1 \(\cos ^{-1}\left(\frac{4}{21}\right)\)
2 \(\sin ^{-1}\left(\frac{4}{21}\right)\)
3 \(\sin ^{-1}\left(\frac{5}{7}\right)\)
4 \(\cos ^{-1}\left(\frac{5}{7}\right)\)
Explanation:
B Angle between line and plane is given by \(\sin \theta=\frac{\left \vert\mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2+\mathrm{c}_1 \mathrm{c}_2\right \vert}{\sqrt{\mathrm{a}_1^2+\mathrm{b}_1^2+\mathrm{c}_1^2} \sqrt{\mathrm{a}_2^2+\mathrm{b}_2^2+\mathrm{c}_2^2}}\)
Given, \(a_1=2, b_1=1, c_1=2\) and
\(\mathrm{a}_2=6, \mathrm{~b}_2=-2, \mathrm{c}_2=-3\)
\(\therefore \sin \theta=\frac{12-2-6}{3 \times 7}=\frac{4}{21} \Rightarrow \theta=\sin ^{-1}\left(\frac{4}{21}\right)\)
MHT CET-2020
Three Dimensional Geometry
121315
If the line \(\frac{x-1}{4}=\frac{y-2}{2}=\frac{z-k}{-2}\) lies in the plane \(2 x+5 y-z=5\), then \(k=\)
1 10
2 7
3 -7
4 -10
Explanation:
B Given, Line \(\frac{x-1}{4}=\frac{y-2}{-2}=\frac{z-k}{-2} \quad\) lies in the plane \(2 x+5 y-z=5\) The plane passes through the point \((1,2, \mathrm{k})\) \(\Rightarrow 2(1)+5(2)+(-1) \mathrm{k}=5\) \(\therefore 2+10-\mathrm{k}=5\) \(\Rightarrow \mathrm{k}=7\)
121309
The angle between the lines \(\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}\) and \(\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-4}{1}\) is
1 \(\cos ^{-1}\left(\frac{1}{3}\right)\)
2 \(\cos ^{-1}\left(\frac{3}{4}\right)\)
3 \(\cos ^{-1}\left(\frac{1}{2}\right)\)
4 \(\cos ^{-1}\left(\frac{2}{3}\right)\)
Explanation:
D Given, Line, \(\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}\) and \(\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-4}{1}\)
\(\therefore \vec{a}=4 \hat{i}+\hat{j}+8 \hat{k} \text { and } \vec{b}=2 \hat{i}+2 \hat{j}+\hat{k}\)
\(\vec{a} \cdot \vec{b}=(4 \hat{i}+\hat{j}+8 \hat{k}) \cdot(2 \hat{i}+2 \hat{j}+\hat{k}) \quad(\because \hat{i} \cdot \hat{i}=\hat{j} \cdot \hat{j}=\hat{k} \cdot \hat{k}=1)\)
\(\vec{a} \cdot \vec{b}=(4 \times 2)+(1 \times 2)+(8 \times 1)=8+2+8=18\)
\(\quad \vert\vec{a} \vert=\sqrt{16+1+64}=\sqrt{81}=9\)
And, \(\quad \vert\vec{b} \vert=\sqrt{4+4+1}=\sqrt{9}=9\)
Let, \(\theta\) be the acute angle between the two given lines.
\(\cos \theta=\frac{\vec{a} \cdot \vec{b}}{ \vert\vec{a} \vert \vert\vec{b} \vert}=\frac{18}{9 \times 3}=\frac{2}{3}\)
\(\theta=\cos ^{-1}\left(\frac{2}{3}\right)\)
MHT CET-2020
Three Dimensional Geometry
121312
The angle between the two lines \(\frac{x-4}{1}=\frac{y+4}{2}=\frac{z+1}{2}\) and \(\frac{x+1}{2}=\frac{y+3}{2}=\frac{z-4}{-1}\) is
1 \(\cos ^{-1}\left(\frac{1}{3}\right)\)
2 \(\cos ^{-1}\left(\frac{4}{9}\right)\)
3 \(\cos ^{-1}\left(\frac{5}{9}\right)\)
4 \(\cos ^{-1}\left(\frac{2}{9}\right)\)
Explanation:
B Given, \(\mathrm{a}_1=2, \quad \mathrm{~b}_1=2 \quad \mathrm{c}_1=-1\)
\(\mathrm{a}_2=1, \quad \mathrm{~b}_2=2, \quad \mathrm{c}_2=2\)
We know that,
\(\cos \theta=\left \vert\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right \vert\)
Let, \(\theta\) be the required angle
\(\cos \theta=\left \vert\frac{1(2)+2(2)+2(-1)}{\sqrt{1+4+4} \sqrt{4+4+1}}\right \vert=\left \vert\frac{2+4-2}{3 \times 3}\right \vert\)
\(\cos \theta=\frac{4}{9}\)
\(\theta=\cos ^{-1}\left(\frac{4}{9}\right)\)
MHT CET-2020
Three Dimensional Geometry
121313
If the angle between the lines whose direction ratio are \(4,-3,5\) and \(3,4, \mathrm{k}\) is \(\frac{\pi}{3}\), then \(\mathrm{k}=\)
1 \(\pm 10\)
2 \(\pm 7\)
3 \(\pm 6\)
4 \(\pm 5\)
Explanation:
D The angle between two line, \(\cos \frac{\pi}{3}=\left \vert\frac{4(3)+(-3)(4)+5 \mathrm{k}}{\sqrt{4^2+(-3)^2+5^2} \sqrt{3^2+4^2+\mathrm{k}^2}}\right \vert\)
\(\therefore \frac{1}{2}=\left \vert\frac{5 \mathrm{k}}{5 \sqrt{2} \sqrt{25+\mathrm{k}^2}}\right \vert\)
On squaring both side we get-
\(\frac{1}{4}=\frac{25 \mathrm{k}^2}{50 \times\left(25+\mathrm{k}^2\right)}\)
\(\therefore 100 \mathrm{k}^2=50\left(25+\mathrm{k}^2\right) \Rightarrow \mathrm{k}^2=25 \Rightarrow \mathrm{k}= \pm 5\)
MHT CET-2020
Three Dimensional Geometry
121314
The angle between the line \(\frac{x-1}{2}=\frac{y+3}{1}=\frac{z+7}{2}\) and the plane \(\overline{\mathbf{r}} \cdot(6 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})=5\) is
1 \(\cos ^{-1}\left(\frac{4}{21}\right)\)
2 \(\sin ^{-1}\left(\frac{4}{21}\right)\)
3 \(\sin ^{-1}\left(\frac{5}{7}\right)\)
4 \(\cos ^{-1}\left(\frac{5}{7}\right)\)
Explanation:
B Angle between line and plane is given by \(\sin \theta=\frac{\left \vert\mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2+\mathrm{c}_1 \mathrm{c}_2\right \vert}{\sqrt{\mathrm{a}_1^2+\mathrm{b}_1^2+\mathrm{c}_1^2} \sqrt{\mathrm{a}_2^2+\mathrm{b}_2^2+\mathrm{c}_2^2}}\)
Given, \(a_1=2, b_1=1, c_1=2\) and
\(\mathrm{a}_2=6, \mathrm{~b}_2=-2, \mathrm{c}_2=-3\)
\(\therefore \sin \theta=\frac{12-2-6}{3 \times 7}=\frac{4}{21} \Rightarrow \theta=\sin ^{-1}\left(\frac{4}{21}\right)\)
MHT CET-2020
Three Dimensional Geometry
121315
If the line \(\frac{x-1}{4}=\frac{y-2}{2}=\frac{z-k}{-2}\) lies in the plane \(2 x+5 y-z=5\), then \(k=\)
1 10
2 7
3 -7
4 -10
Explanation:
B Given, Line \(\frac{x-1}{4}=\frac{y-2}{-2}=\frac{z-k}{-2} \quad\) lies in the plane \(2 x+5 y-z=5\) The plane passes through the point \((1,2, \mathrm{k})\) \(\Rightarrow 2(1)+5(2)+(-1) \mathrm{k}=5\) \(\therefore 2+10-\mathrm{k}=5\) \(\Rightarrow \mathrm{k}=7\)
121309
The angle between the lines \(\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}\) and \(\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-4}{1}\) is
1 \(\cos ^{-1}\left(\frac{1}{3}\right)\)
2 \(\cos ^{-1}\left(\frac{3}{4}\right)\)
3 \(\cos ^{-1}\left(\frac{1}{2}\right)\)
4 \(\cos ^{-1}\left(\frac{2}{3}\right)\)
Explanation:
D Given, Line, \(\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}\) and \(\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-4}{1}\)
\(\therefore \vec{a}=4 \hat{i}+\hat{j}+8 \hat{k} \text { and } \vec{b}=2 \hat{i}+2 \hat{j}+\hat{k}\)
\(\vec{a} \cdot \vec{b}=(4 \hat{i}+\hat{j}+8 \hat{k}) \cdot(2 \hat{i}+2 \hat{j}+\hat{k}) \quad(\because \hat{i} \cdot \hat{i}=\hat{j} \cdot \hat{j}=\hat{k} \cdot \hat{k}=1)\)
\(\vec{a} \cdot \vec{b}=(4 \times 2)+(1 \times 2)+(8 \times 1)=8+2+8=18\)
\(\quad \vert\vec{a} \vert=\sqrt{16+1+64}=\sqrt{81}=9\)
And, \(\quad \vert\vec{b} \vert=\sqrt{4+4+1}=\sqrt{9}=9\)
Let, \(\theta\) be the acute angle between the two given lines.
\(\cos \theta=\frac{\vec{a} \cdot \vec{b}}{ \vert\vec{a} \vert \vert\vec{b} \vert}=\frac{18}{9 \times 3}=\frac{2}{3}\)
\(\theta=\cos ^{-1}\left(\frac{2}{3}\right)\)
MHT CET-2020
Three Dimensional Geometry
121312
The angle between the two lines \(\frac{x-4}{1}=\frac{y+4}{2}=\frac{z+1}{2}\) and \(\frac{x+1}{2}=\frac{y+3}{2}=\frac{z-4}{-1}\) is
1 \(\cos ^{-1}\left(\frac{1}{3}\right)\)
2 \(\cos ^{-1}\left(\frac{4}{9}\right)\)
3 \(\cos ^{-1}\left(\frac{5}{9}\right)\)
4 \(\cos ^{-1}\left(\frac{2}{9}\right)\)
Explanation:
B Given, \(\mathrm{a}_1=2, \quad \mathrm{~b}_1=2 \quad \mathrm{c}_1=-1\)
\(\mathrm{a}_2=1, \quad \mathrm{~b}_2=2, \quad \mathrm{c}_2=2\)
We know that,
\(\cos \theta=\left \vert\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right \vert\)
Let, \(\theta\) be the required angle
\(\cos \theta=\left \vert\frac{1(2)+2(2)+2(-1)}{\sqrt{1+4+4} \sqrt{4+4+1}}\right \vert=\left \vert\frac{2+4-2}{3 \times 3}\right \vert\)
\(\cos \theta=\frac{4}{9}\)
\(\theta=\cos ^{-1}\left(\frac{4}{9}\right)\)
MHT CET-2020
Three Dimensional Geometry
121313
If the angle between the lines whose direction ratio are \(4,-3,5\) and \(3,4, \mathrm{k}\) is \(\frac{\pi}{3}\), then \(\mathrm{k}=\)
1 \(\pm 10\)
2 \(\pm 7\)
3 \(\pm 6\)
4 \(\pm 5\)
Explanation:
D The angle between two line, \(\cos \frac{\pi}{3}=\left \vert\frac{4(3)+(-3)(4)+5 \mathrm{k}}{\sqrt{4^2+(-3)^2+5^2} \sqrt{3^2+4^2+\mathrm{k}^2}}\right \vert\)
\(\therefore \frac{1}{2}=\left \vert\frac{5 \mathrm{k}}{5 \sqrt{2} \sqrt{25+\mathrm{k}^2}}\right \vert\)
On squaring both side we get-
\(\frac{1}{4}=\frac{25 \mathrm{k}^2}{50 \times\left(25+\mathrm{k}^2\right)}\)
\(\therefore 100 \mathrm{k}^2=50\left(25+\mathrm{k}^2\right) \Rightarrow \mathrm{k}^2=25 \Rightarrow \mathrm{k}= \pm 5\)
MHT CET-2020
Three Dimensional Geometry
121314
The angle between the line \(\frac{x-1}{2}=\frac{y+3}{1}=\frac{z+7}{2}\) and the plane \(\overline{\mathbf{r}} \cdot(6 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})=5\) is
1 \(\cos ^{-1}\left(\frac{4}{21}\right)\)
2 \(\sin ^{-1}\left(\frac{4}{21}\right)\)
3 \(\sin ^{-1}\left(\frac{5}{7}\right)\)
4 \(\cos ^{-1}\left(\frac{5}{7}\right)\)
Explanation:
B Angle between line and plane is given by \(\sin \theta=\frac{\left \vert\mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2+\mathrm{c}_1 \mathrm{c}_2\right \vert}{\sqrt{\mathrm{a}_1^2+\mathrm{b}_1^2+\mathrm{c}_1^2} \sqrt{\mathrm{a}_2^2+\mathrm{b}_2^2+\mathrm{c}_2^2}}\)
Given, \(a_1=2, b_1=1, c_1=2\) and
\(\mathrm{a}_2=6, \mathrm{~b}_2=-2, \mathrm{c}_2=-3\)
\(\therefore \sin \theta=\frac{12-2-6}{3 \times 7}=\frac{4}{21} \Rightarrow \theta=\sin ^{-1}\left(\frac{4}{21}\right)\)
MHT CET-2020
Three Dimensional Geometry
121315
If the line \(\frac{x-1}{4}=\frac{y-2}{2}=\frac{z-k}{-2}\) lies in the plane \(2 x+5 y-z=5\), then \(k=\)
1 10
2 7
3 -7
4 -10
Explanation:
B Given, Line \(\frac{x-1}{4}=\frac{y-2}{-2}=\frac{z-k}{-2} \quad\) lies in the plane \(2 x+5 y-z=5\) The plane passes through the point \((1,2, \mathrm{k})\) \(\Rightarrow 2(1)+5(2)+(-1) \mathrm{k}=5\) \(\therefore 2+10-\mathrm{k}=5\) \(\Rightarrow \mathrm{k}=7\)
121309
The angle between the lines \(\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}\) and \(\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-4}{1}\) is
1 \(\cos ^{-1}\left(\frac{1}{3}\right)\)
2 \(\cos ^{-1}\left(\frac{3}{4}\right)\)
3 \(\cos ^{-1}\left(\frac{1}{2}\right)\)
4 \(\cos ^{-1}\left(\frac{2}{3}\right)\)
Explanation:
D Given, Line, \(\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}\) and \(\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-4}{1}\)
\(\therefore \vec{a}=4 \hat{i}+\hat{j}+8 \hat{k} \text { and } \vec{b}=2 \hat{i}+2 \hat{j}+\hat{k}\)
\(\vec{a} \cdot \vec{b}=(4 \hat{i}+\hat{j}+8 \hat{k}) \cdot(2 \hat{i}+2 \hat{j}+\hat{k}) \quad(\because \hat{i} \cdot \hat{i}=\hat{j} \cdot \hat{j}=\hat{k} \cdot \hat{k}=1)\)
\(\vec{a} \cdot \vec{b}=(4 \times 2)+(1 \times 2)+(8 \times 1)=8+2+8=18\)
\(\quad \vert\vec{a} \vert=\sqrt{16+1+64}=\sqrt{81}=9\)
And, \(\quad \vert\vec{b} \vert=\sqrt{4+4+1}=\sqrt{9}=9\)
Let, \(\theta\) be the acute angle between the two given lines.
\(\cos \theta=\frac{\vec{a} \cdot \vec{b}}{ \vert\vec{a} \vert \vert\vec{b} \vert}=\frac{18}{9 \times 3}=\frac{2}{3}\)
\(\theta=\cos ^{-1}\left(\frac{2}{3}\right)\)
MHT CET-2020
Three Dimensional Geometry
121312
The angle between the two lines \(\frac{x-4}{1}=\frac{y+4}{2}=\frac{z+1}{2}\) and \(\frac{x+1}{2}=\frac{y+3}{2}=\frac{z-4}{-1}\) is
1 \(\cos ^{-1}\left(\frac{1}{3}\right)\)
2 \(\cos ^{-1}\left(\frac{4}{9}\right)\)
3 \(\cos ^{-1}\left(\frac{5}{9}\right)\)
4 \(\cos ^{-1}\left(\frac{2}{9}\right)\)
Explanation:
B Given, \(\mathrm{a}_1=2, \quad \mathrm{~b}_1=2 \quad \mathrm{c}_1=-1\)
\(\mathrm{a}_2=1, \quad \mathrm{~b}_2=2, \quad \mathrm{c}_2=2\)
We know that,
\(\cos \theta=\left \vert\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right \vert\)
Let, \(\theta\) be the required angle
\(\cos \theta=\left \vert\frac{1(2)+2(2)+2(-1)}{\sqrt{1+4+4} \sqrt{4+4+1}}\right \vert=\left \vert\frac{2+4-2}{3 \times 3}\right \vert\)
\(\cos \theta=\frac{4}{9}\)
\(\theta=\cos ^{-1}\left(\frac{4}{9}\right)\)
MHT CET-2020
Three Dimensional Geometry
121313
If the angle between the lines whose direction ratio are \(4,-3,5\) and \(3,4, \mathrm{k}\) is \(\frac{\pi}{3}\), then \(\mathrm{k}=\)
1 \(\pm 10\)
2 \(\pm 7\)
3 \(\pm 6\)
4 \(\pm 5\)
Explanation:
D The angle between two line, \(\cos \frac{\pi}{3}=\left \vert\frac{4(3)+(-3)(4)+5 \mathrm{k}}{\sqrt{4^2+(-3)^2+5^2} \sqrt{3^2+4^2+\mathrm{k}^2}}\right \vert\)
\(\therefore \frac{1}{2}=\left \vert\frac{5 \mathrm{k}}{5 \sqrt{2} \sqrt{25+\mathrm{k}^2}}\right \vert\)
On squaring both side we get-
\(\frac{1}{4}=\frac{25 \mathrm{k}^2}{50 \times\left(25+\mathrm{k}^2\right)}\)
\(\therefore 100 \mathrm{k}^2=50\left(25+\mathrm{k}^2\right) \Rightarrow \mathrm{k}^2=25 \Rightarrow \mathrm{k}= \pm 5\)
MHT CET-2020
Three Dimensional Geometry
121314
The angle between the line \(\frac{x-1}{2}=\frac{y+3}{1}=\frac{z+7}{2}\) and the plane \(\overline{\mathbf{r}} \cdot(6 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})=5\) is
1 \(\cos ^{-1}\left(\frac{4}{21}\right)\)
2 \(\sin ^{-1}\left(\frac{4}{21}\right)\)
3 \(\sin ^{-1}\left(\frac{5}{7}\right)\)
4 \(\cos ^{-1}\left(\frac{5}{7}\right)\)
Explanation:
B Angle between line and plane is given by \(\sin \theta=\frac{\left \vert\mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2+\mathrm{c}_1 \mathrm{c}_2\right \vert}{\sqrt{\mathrm{a}_1^2+\mathrm{b}_1^2+\mathrm{c}_1^2} \sqrt{\mathrm{a}_2^2+\mathrm{b}_2^2+\mathrm{c}_2^2}}\)
Given, \(a_1=2, b_1=1, c_1=2\) and
\(\mathrm{a}_2=6, \mathrm{~b}_2=-2, \mathrm{c}_2=-3\)
\(\therefore \sin \theta=\frac{12-2-6}{3 \times 7}=\frac{4}{21} \Rightarrow \theta=\sin ^{-1}\left(\frac{4}{21}\right)\)
MHT CET-2020
Three Dimensional Geometry
121315
If the line \(\frac{x-1}{4}=\frac{y-2}{2}=\frac{z-k}{-2}\) lies in the plane \(2 x+5 y-z=5\), then \(k=\)
1 10
2 7
3 -7
4 -10
Explanation:
B Given, Line \(\frac{x-1}{4}=\frac{y-2}{-2}=\frac{z-k}{-2} \quad\) lies in the plane \(2 x+5 y-z=5\) The plane passes through the point \((1,2, \mathrm{k})\) \(\Rightarrow 2(1)+5(2)+(-1) \mathrm{k}=5\) \(\therefore 2+10-\mathrm{k}=5\) \(\Rightarrow \mathrm{k}=7\)
121309
The angle between the lines \(\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}\) and \(\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-4}{1}\) is
1 \(\cos ^{-1}\left(\frac{1}{3}\right)\)
2 \(\cos ^{-1}\left(\frac{3}{4}\right)\)
3 \(\cos ^{-1}\left(\frac{1}{2}\right)\)
4 \(\cos ^{-1}\left(\frac{2}{3}\right)\)
Explanation:
D Given, Line, \(\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}\) and \(\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-4}{1}\)
\(\therefore \vec{a}=4 \hat{i}+\hat{j}+8 \hat{k} \text { and } \vec{b}=2 \hat{i}+2 \hat{j}+\hat{k}\)
\(\vec{a} \cdot \vec{b}=(4 \hat{i}+\hat{j}+8 \hat{k}) \cdot(2 \hat{i}+2 \hat{j}+\hat{k}) \quad(\because \hat{i} \cdot \hat{i}=\hat{j} \cdot \hat{j}=\hat{k} \cdot \hat{k}=1)\)
\(\vec{a} \cdot \vec{b}=(4 \times 2)+(1 \times 2)+(8 \times 1)=8+2+8=18\)
\(\quad \vert\vec{a} \vert=\sqrt{16+1+64}=\sqrt{81}=9\)
And, \(\quad \vert\vec{b} \vert=\sqrt{4+4+1}=\sqrt{9}=9\)
Let, \(\theta\) be the acute angle between the two given lines.
\(\cos \theta=\frac{\vec{a} \cdot \vec{b}}{ \vert\vec{a} \vert \vert\vec{b} \vert}=\frac{18}{9 \times 3}=\frac{2}{3}\)
\(\theta=\cos ^{-1}\left(\frac{2}{3}\right)\)
MHT CET-2020
Three Dimensional Geometry
121312
The angle between the two lines \(\frac{x-4}{1}=\frac{y+4}{2}=\frac{z+1}{2}\) and \(\frac{x+1}{2}=\frac{y+3}{2}=\frac{z-4}{-1}\) is
1 \(\cos ^{-1}\left(\frac{1}{3}\right)\)
2 \(\cos ^{-1}\left(\frac{4}{9}\right)\)
3 \(\cos ^{-1}\left(\frac{5}{9}\right)\)
4 \(\cos ^{-1}\left(\frac{2}{9}\right)\)
Explanation:
B Given, \(\mathrm{a}_1=2, \quad \mathrm{~b}_1=2 \quad \mathrm{c}_1=-1\)
\(\mathrm{a}_2=1, \quad \mathrm{~b}_2=2, \quad \mathrm{c}_2=2\)
We know that,
\(\cos \theta=\left \vert\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right \vert\)
Let, \(\theta\) be the required angle
\(\cos \theta=\left \vert\frac{1(2)+2(2)+2(-1)}{\sqrt{1+4+4} \sqrt{4+4+1}}\right \vert=\left \vert\frac{2+4-2}{3 \times 3}\right \vert\)
\(\cos \theta=\frac{4}{9}\)
\(\theta=\cos ^{-1}\left(\frac{4}{9}\right)\)
MHT CET-2020
Three Dimensional Geometry
121313
If the angle between the lines whose direction ratio are \(4,-3,5\) and \(3,4, \mathrm{k}\) is \(\frac{\pi}{3}\), then \(\mathrm{k}=\)
1 \(\pm 10\)
2 \(\pm 7\)
3 \(\pm 6\)
4 \(\pm 5\)
Explanation:
D The angle between two line, \(\cos \frac{\pi}{3}=\left \vert\frac{4(3)+(-3)(4)+5 \mathrm{k}}{\sqrt{4^2+(-3)^2+5^2} \sqrt{3^2+4^2+\mathrm{k}^2}}\right \vert\)
\(\therefore \frac{1}{2}=\left \vert\frac{5 \mathrm{k}}{5 \sqrt{2} \sqrt{25+\mathrm{k}^2}}\right \vert\)
On squaring both side we get-
\(\frac{1}{4}=\frac{25 \mathrm{k}^2}{50 \times\left(25+\mathrm{k}^2\right)}\)
\(\therefore 100 \mathrm{k}^2=50\left(25+\mathrm{k}^2\right) \Rightarrow \mathrm{k}^2=25 \Rightarrow \mathrm{k}= \pm 5\)
MHT CET-2020
Three Dimensional Geometry
121314
The angle between the line \(\frac{x-1}{2}=\frac{y+3}{1}=\frac{z+7}{2}\) and the plane \(\overline{\mathbf{r}} \cdot(6 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})=5\) is
1 \(\cos ^{-1}\left(\frac{4}{21}\right)\)
2 \(\sin ^{-1}\left(\frac{4}{21}\right)\)
3 \(\sin ^{-1}\left(\frac{5}{7}\right)\)
4 \(\cos ^{-1}\left(\frac{5}{7}\right)\)
Explanation:
B Angle between line and plane is given by \(\sin \theta=\frac{\left \vert\mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2+\mathrm{c}_1 \mathrm{c}_2\right \vert}{\sqrt{\mathrm{a}_1^2+\mathrm{b}_1^2+\mathrm{c}_1^2} \sqrt{\mathrm{a}_2^2+\mathrm{b}_2^2+\mathrm{c}_2^2}}\)
Given, \(a_1=2, b_1=1, c_1=2\) and
\(\mathrm{a}_2=6, \mathrm{~b}_2=-2, \mathrm{c}_2=-3\)
\(\therefore \sin \theta=\frac{12-2-6}{3 \times 7}=\frac{4}{21} \Rightarrow \theta=\sin ^{-1}\left(\frac{4}{21}\right)\)
MHT CET-2020
Three Dimensional Geometry
121315
If the line \(\frac{x-1}{4}=\frac{y-2}{2}=\frac{z-k}{-2}\) lies in the plane \(2 x+5 y-z=5\), then \(k=\)
1 10
2 7
3 -7
4 -10
Explanation:
B Given, Line \(\frac{x-1}{4}=\frac{y-2}{-2}=\frac{z-k}{-2} \quad\) lies in the plane \(2 x+5 y-z=5\) The plane passes through the point \((1,2, \mathrm{k})\) \(\Rightarrow 2(1)+5(2)+(-1) \mathrm{k}=5\) \(\therefore 2+10-\mathrm{k}=5\) \(\Rightarrow \mathrm{k}=7\)
