Top 50 Most Repeated THREE DIMENSIONAL GEOMETRY PYQs | JEE MAINS
A curated collection of the most important questions from THREE DIMENSIONAL GEOMETRY, fully solved with step-by-step concepts to prepare for JEE MAINS.
A curated collection of the most important questions from THREE DIMENSIONAL GEOMETRY, fully solved with step-by-step concepts to prepare for JEE MAINS.
Use distance formula: $\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}} = \frac{|2(1)-3(-2)+6(3)+14|}{\sqrt{4+9+36}} = \frac{42}{7} = 6$. Wait: recalcula...
Read Full Step-by-Step Solution →Step 1: Vector $\overrightarrow{PQ}= (4-1,6-2,9-3) = (3,4,6)$. Hence direction ratios are $3:4:6$. Step 2: Magnitude $|\overrightarrow{PQ}|=\sqrt{3^2+...
Read Full Step-by-Step Solution →Direction cosines are the direction ratios divided by their magnitude. \(\sqrt{2^2+(-1)^2+2^2}=3\), so the cosines are \(2/3, -1/3, 2/3\)....
Read Full Step-by-Step Solution →Use $\cos\theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$....
Read Full Step-by-Step Solution →Use distance formula: $\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}$....
Read Full Step-by-Step Solution →Symmetric form: $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ for point $(x_1,y_1,z_1)$ and direction ratios $a,b,c$....
Read Full Step-by-Step Solution →The normals are \(\mathbf{n}_1=(2,-1,2)\) and \(\mathbf{n}_2=(1,2,-2)\). Using \(\cos\theta = \frac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{|\mathbf{n}_1||\m...
Read Full Step-by-Step Solution →Using $\vec{n}\cdot(\vec{r}-\vec{r_0})=0$ gives $x+2y-2z+6=0$ or $x+2y-2z=-6$. Setting two coordinates to zero yields the intercepts $a=-6$, $b=-3$, $...
Read Full Step-by-Step Solution →If two lines are parallel, their dot product is zero....
Read Full Step-by-Step Solution →Normal form: $ \vec{r} \cdot \hat{n} = d $. Here $ |\vec{n}| = 3 $, so $ \vec{r} \cdot \vec{n} = 9 $....
Read Full Step-by-Step Solution →Using direction cosines identity $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$....
Read Full Step-by-Step Solution →Using $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$, so $\cos^2\gamma = 1 - 0.75 = 0.25$...
Read Full Step-by-Step Solution →Using $\cos\theta=\dfrac{\mathbf{d}_1\cdot\mathbf{d}_2}{|\mathbf{d}_1|\,|\mathbf{d}_2|}$, we get $\mathbf{d}_1\cdot\mathbf{d}_2=4$, $|\mathbf{d}_1|=|\...
Read Full Step-by-Step Solution →The intercepts are directly read from the points: $a=2$, $b=3$, $c=4$. Their product $abc = 2\times3\times4 = 24$....
Read Full Step-by-Step Solution →Write the line in parametric form: $x=1+2t$, $y=-3-t$, $z=4+3t$. Substitute into the plane equation:
Intercepts are all 6. Sum of reciprocals = 1/6 + 1/6 + 1/6 = 1/2 × 3 = 0.5 → Wait: 3/6 = 0.5? No: 1/6+1/6+1/6=1/2? No: 3/6=0.5 → correction: 3/6 = 0.5...
Read Full Step-by-Step Solution →For coplanar lines, scalar triple product of direction vectors and position vector difference is zero. Here, solving $\vec{d_1} \cdot (\vec{d_2} \time...
Read Full Step-by-Step Solution →Foundational check for Three Dimensional Geometry in Class 12. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Three Dimensional Ge...
Read Full Step-by-Step Solution →Use distance formula: $\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}} = \frac{|2(1)-3(-2)+6(3)+14|}{\sqrt{4+9+36}} = \frac{42}{7} = 6$. Wait: recalcula...
Read Full Step-by-Step Solution →Step 1: Vector $\overrightarrow{PQ}= (4-1,6-2,9-3) = (3,4,6)$. Hence direction ratios are $3:4:6$. Step 2: Magnitude $|\overrightarrow{PQ}|=\sqrt{3^2+...
Read Full Step-by-Step Solution →Direction cosines are the direction ratios divided by their magnitude. \(\sqrt{2^2+(-1)^2+2^2}=3\), so the cosines are \(2/3, -1/3, 2/3\)....
Read Full Step-by-Step Solution →Use $\cos\theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$....
Read Full Step-by-Step Solution →Use distance formula: $\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}$....
Read Full Step-by-Step Solution →Symmetric form: $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ for point $(x_1,y_1,z_1)$ and direction ratios $a,b,c$....
Read Full Step-by-Step Solution →The normals are \(\mathbf{n}_1=(2,-1,2)\) and \(\mathbf{n}_2=(1,2,-2)\). Using \(\cos\theta = \frac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{|\mathbf{n}_1||\m...
Read Full Step-by-Step Solution →Using $\vec{n}\cdot(\vec{r}-\vec{r_0})=0$ gives $x+2y-2z+6=0$ or $x+2y-2z=-6$. Setting two coordinates to zero yields the intercepts $a=-6$, $b=-3$, $...
Read Full Step-by-Step Solution →If two lines are parallel, their dot product is zero....
Read Full Step-by-Step Solution →Normal form: $ \vec{r} \cdot \hat{n} = d $. Here $ |\vec{n}| = 3 $, so $ \vec{r} \cdot \vec{n} = 9 $....
Read Full Step-by-Step Solution →Using direction cosines identity $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$....
Read Full Step-by-Step Solution →Using $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$, so $\cos^2\gamma = 1 - 0.75 = 0.25$...
Read Full Step-by-Step Solution →Using $\cos\theta=\dfrac{\mathbf{d}_1\cdot\mathbf{d}_2}{|\mathbf{d}_1|\,|\mathbf{d}_2|}$, we get $\mathbf{d}_1\cdot\mathbf{d}_2=4$, $|\mathbf{d}_1|=|\...
Read Full Step-by-Step Solution →The intercepts are directly read from the points: $a=2$, $b=3$, $c=4$. Their product $abc = 2\times3\times4 = 24$....
Read Full Step-by-Step Solution →Write the line in parametric form: $x=1+2t$, $y=-3-t$, $z=4+3t$. Substitute into the plane equation:
Intercepts are all 6. Sum of reciprocals = 1/6 + 1/6 + 1/6 = 1/2 × 3 = 0.5 → Wait: 3/6 = 0.5? No: 1/6+1/6+1/6=1/2? No: 3/6=0.5 → correction: 3/6 = 0.5...
Read Full Step-by-Step Solution →For coplanar lines, scalar triple product of direction vectors and position vector difference is zero. Here, solving $\vec{d_1} \cdot (\vec{d_2} \time...
Read Full Step-by-Step Solution →Foundational check for Three Dimensional Geometry in Class 12. Study the core principles carefully for competitive exams....
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Three Dimensional Ge...
Read Full Step-by-Step Solution →Use distance formula: $\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}} = \frac{|2(1)-3(-2)+6(3)+14|}{\sqrt{4+9+36}} = \frac{42}{7} = 6$. Wait: recalcula...
Read Full Step-by-Step Solution →Step 1: Vector $\overrightarrow{PQ}= (4-1,6-2,9-3) = (3,4,6)$. Hence direction ratios are $3:4:6$. Step 2: Magnitude $|\overrightarrow{PQ}|=\sqrt{3^2+...
Read Full Step-by-Step Solution →Direction cosines are the direction ratios divided by their magnitude. \(\sqrt{2^2+(-1)^2+2^2}=3\), so the cosines are \(2/3, -1/3, 2/3\)....
Read Full Step-by-Step Solution →Use $\cos\theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$....
Read Full Step-by-Step Solution →Use distance formula: $\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}$....
Read Full Step-by-Step Solution →Symmetric form: $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ for point $(x_1,y_1,z_1)$ and direction ratios $a,b,c$....
Read Full Step-by-Step Solution →The normals are \(\mathbf{n}_1=(2,-1,2)\) and \(\mathbf{n}_2=(1,2,-2)\). Using \(\cos\theta = \frac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{|\mathbf{n}_1||\m...
Read Full Step-by-Step Solution →Using $\vec{n}\cdot(\vec{r}-\vec{r_0})=0$ gives $x+2y-2z+6=0$ or $x+2y-2z=-6$. Setting two coordinates to zero yields the intercepts $a=-6$, $b=-3$, $...
Read Full Step-by-Step Solution →If two lines are parallel, their dot product is zero....
Read Full Step-by-Step Solution →Normal form: $ \vec{r} \cdot \hat{n} = d $. Here $ |\vec{n}| = 3 $, so $ \vec{r} \cdot \vec{n} = 9 $....
Read Full Step-by-Step Solution →Using direction cosines identity $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$....
Read Full Step-by-Step Solution →Using $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$, so $\cos^2\gamma = 1 - 0.75 = 0.25$...
Read Full Step-by-Step Solution →