Cartesian and Vector Equations of Line and Plane
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Three Dimensional Geometry

121194 If Cartesian equation of the line is \(x-1=2 y+\) \(3=3-z\), then its vector equation is

1 \(\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})\)
2 \(\overrightarrow{\mathrm{r}}=(-\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda\left(\hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}-\hat{\mathrm{k}}\right)\)
3 \(\overrightarrow{\mathrm{r}}=\left(\hat{\mathrm{i}}-\frac{3}{2} \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\right)+\lambda(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})\)
4 \(\overrightarrow{\mathrm{r}}=\left(-\hat{\mathrm{i}}+\frac{3}{2} \hat{\mathrm{j}}-3 \hat{\mathrm{k}}\right)+\lambda(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})\)
MHT CET-2020
Three Dimensional Geometry

121195 The equation of the line passing through the point \((1,2,3)\) and perpendicular to the lines \(\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}\) and \(\overline{\mathbf{r}}=\lambda(-3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})\) is

1 \(\bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}-7 \hat{j}+4 \hat{k})\)
2 \(\bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}+7 \hat{j}-4 \hat{k})\)
3 \(\bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}-7 \hat{j}-4 \hat{k})\)
4 \(\bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}+7 \hat{j}+4 \hat{k})\)
MHT CET-2020
Three Dimensional Geometry

121198 The equation of the plane passing through the point \((-1,2,1)\) and perpendicular to the line joining the points \((-3,1,2)\) and \((2,3,4)\) is

1 \(\overrightarrow{\mathrm{r}}(5 \hat{\mathrm{i}}-2 \hat{\mathrm{i}}-2 \hat{\mathrm{k}})=1\)
2 \(\overrightarrow{\mathrm{r}}(5 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})=1\)
3 \(\overrightarrow{\mathrm{r}}(5 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})=-1\)
4 \(\overrightarrow{\mathrm{r}}(5 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})=-5\)
MHT CET-2019
Three Dimensional Geometry

121199 If planes \(\overrightarrow{\mathbf{r}} \times(\mathbf{p} \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})+\mathbf{3}=\mathbf{0}\) and \(\overrightarrow{\mathbf{r}} \times(\hat{\mathbf{i}}-\mathbf{p} \hat{\mathbf{j}}-\hat{\mathbf{k}})-5=0\) include angle \(\frac{\pi}{3}\), then the value of \(p\) is

1 \(1,-3\)
2 \(-1,3\)
3 -3
4 3
MHT CET-2018
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Three Dimensional Geometry

121194 If Cartesian equation of the line is \(x-1=2 y+\) \(3=3-z\), then its vector equation is

1 \(\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})\)
2 \(\overrightarrow{\mathrm{r}}=(-\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda\left(\hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}-\hat{\mathrm{k}}\right)\)
3 \(\overrightarrow{\mathrm{r}}=\left(\hat{\mathrm{i}}-\frac{3}{2} \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\right)+\lambda(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})\)
4 \(\overrightarrow{\mathrm{r}}=\left(-\hat{\mathrm{i}}+\frac{3}{2} \hat{\mathrm{j}}-3 \hat{\mathrm{k}}\right)+\lambda(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})\)
MHT CET-2020
Three Dimensional Geometry

121195 The equation of the line passing through the point \((1,2,3)\) and perpendicular to the lines \(\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}\) and \(\overline{\mathbf{r}}=\lambda(-3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})\) is

1 \(\bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}-7 \hat{j}+4 \hat{k})\)
2 \(\bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}+7 \hat{j}-4 \hat{k})\)
3 \(\bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}-7 \hat{j}-4 \hat{k})\)
4 \(\bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}+7 \hat{j}+4 \hat{k})\)
MHT CET-2020
Three Dimensional Geometry

121198 The equation of the plane passing through the point \((-1,2,1)\) and perpendicular to the line joining the points \((-3,1,2)\) and \((2,3,4)\) is

1 \(\overrightarrow{\mathrm{r}}(5 \hat{\mathrm{i}}-2 \hat{\mathrm{i}}-2 \hat{\mathrm{k}})=1\)
2 \(\overrightarrow{\mathrm{r}}(5 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})=1\)
3 \(\overrightarrow{\mathrm{r}}(5 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})=-1\)
4 \(\overrightarrow{\mathrm{r}}(5 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})=-5\)
MHT CET-2019
Three Dimensional Geometry

121199 If planes \(\overrightarrow{\mathbf{r}} \times(\mathbf{p} \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})+\mathbf{3}=\mathbf{0}\) and \(\overrightarrow{\mathbf{r}} \times(\hat{\mathbf{i}}-\mathbf{p} \hat{\mathbf{j}}-\hat{\mathbf{k}})-5=0\) include angle \(\frac{\pi}{3}\), then the value of \(p\) is

1 \(1,-3\)
2 \(-1,3\)
3 -3
4 3
MHT CET-2018
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Three Dimensional Geometry

121194 If Cartesian equation of the line is \(x-1=2 y+\) \(3=3-z\), then its vector equation is

1 \(\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})\)
2 \(\overrightarrow{\mathrm{r}}=(-\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda\left(\hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}-\hat{\mathrm{k}}\right)\)
3 \(\overrightarrow{\mathrm{r}}=\left(\hat{\mathrm{i}}-\frac{3}{2} \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\right)+\lambda(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})\)
4 \(\overrightarrow{\mathrm{r}}=\left(-\hat{\mathrm{i}}+\frac{3}{2} \hat{\mathrm{j}}-3 \hat{\mathrm{k}}\right)+\lambda(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})\)
MHT CET-2020
Three Dimensional Geometry

121195 The equation of the line passing through the point \((1,2,3)\) and perpendicular to the lines \(\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}\) and \(\overline{\mathbf{r}}=\lambda(-3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})\) is

1 \(\bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}-7 \hat{j}+4 \hat{k})\)
2 \(\bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}+7 \hat{j}-4 \hat{k})\)
3 \(\bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}-7 \hat{j}-4 \hat{k})\)
4 \(\bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}+7 \hat{j}+4 \hat{k})\)
MHT CET-2020
Three Dimensional Geometry

121198 The equation of the plane passing through the point \((-1,2,1)\) and perpendicular to the line joining the points \((-3,1,2)\) and \((2,3,4)\) is

1 \(\overrightarrow{\mathrm{r}}(5 \hat{\mathrm{i}}-2 \hat{\mathrm{i}}-2 \hat{\mathrm{k}})=1\)
2 \(\overrightarrow{\mathrm{r}}(5 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})=1\)
3 \(\overrightarrow{\mathrm{r}}(5 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})=-1\)
4 \(\overrightarrow{\mathrm{r}}(5 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})=-5\)
MHT CET-2019
Three Dimensional Geometry

121199 If planes \(\overrightarrow{\mathbf{r}} \times(\mathbf{p} \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})+\mathbf{3}=\mathbf{0}\) and \(\overrightarrow{\mathbf{r}} \times(\hat{\mathbf{i}}-\mathbf{p} \hat{\mathbf{j}}-\hat{\mathbf{k}})-5=0\) include angle \(\frac{\pi}{3}\), then the value of \(p\) is

1 \(1,-3\)
2 \(-1,3\)
3 -3
4 3
MHT CET-2018
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Three Dimensional Geometry

121194 If Cartesian equation of the line is \(x-1=2 y+\) \(3=3-z\), then its vector equation is

1 \(\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})\)
2 \(\overrightarrow{\mathrm{r}}=(-\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda\left(\hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}-\hat{\mathrm{k}}\right)\)
3 \(\overrightarrow{\mathrm{r}}=\left(\hat{\mathrm{i}}-\frac{3}{2} \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\right)+\lambda(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})\)
4 \(\overrightarrow{\mathrm{r}}=\left(-\hat{\mathrm{i}}+\frac{3}{2} \hat{\mathrm{j}}-3 \hat{\mathrm{k}}\right)+\lambda(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})\)
MHT CET-2020
Three Dimensional Geometry

121195 The equation of the line passing through the point \((1,2,3)\) and perpendicular to the lines \(\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}\) and \(\overline{\mathbf{r}}=\lambda(-3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})\) is

1 \(\bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}-7 \hat{j}+4 \hat{k})\)
2 \(\bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}+7 \hat{j}-4 \hat{k})\)
3 \(\bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}-7 \hat{j}-4 \hat{k})\)
4 \(\bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}+7 \hat{j}+4 \hat{k})\)
MHT CET-2020
Three Dimensional Geometry

121198 The equation of the plane passing through the point \((-1,2,1)\) and perpendicular to the line joining the points \((-3,1,2)\) and \((2,3,4)\) is

1 \(\overrightarrow{\mathrm{r}}(5 \hat{\mathrm{i}}-2 \hat{\mathrm{i}}-2 \hat{\mathrm{k}})=1\)
2 \(\overrightarrow{\mathrm{r}}(5 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})=1\)
3 \(\overrightarrow{\mathrm{r}}(5 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})=-1\)
4 \(\overrightarrow{\mathrm{r}}(5 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})=-5\)
MHT CET-2019
Three Dimensional Geometry

121199 If planes \(\overrightarrow{\mathbf{r}} \times(\mathbf{p} \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})+\mathbf{3}=\mathbf{0}\) and \(\overrightarrow{\mathbf{r}} \times(\hat{\mathbf{i}}-\mathbf{p} \hat{\mathbf{j}}-\hat{\mathbf{k}})-5=0\) include angle \(\frac{\pi}{3}\), then the value of \(p\) is

1 \(1,-3\)
2 \(-1,3\)
3 -3
4 3
MHT CET-2018
For More Updates join with Us