NCERT Class 12 Maths Misc. Ex. Ch-11 (3D Geometry)

The solutions to the NCERT Class 12 Mathematics Miscellaneous Exercise for Chapter 11, focusing on Three Dimensional Geometry, constitute an indispensable resource for students seeking to master this complex subject. This particular exercise is meticulously designed to consolidate and extend the understanding of advanced topics, including the vector and Cartesian equations of lines, the shortest distance between skew lines, the angle between two lines, and questions related to coplanarity. Working through these problems, often characterized by their multi-conceptual nature, demands a robust grasp of foundational principles and analytical reasoning; comprehensive solutions offer invaluable step-by-step guidance, elucidating the correct application of formulae, logical progression of steps, and effective problem-solving strategies. Consequently, a thorough engagement with these detailed solutions is paramount for students aiming to solidify their conceptual understanding, enhance their problem-solving acumen, and achieve academic excellence in Three Dimensional Geometry. Get Free NCERT Class 12 Maths Misc. Ex. Ch-11 (3D Geometry) solution here.

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Q.1. NCERT Class 12 Maths Misc. Ex. Ch-11 (3D Geometry) Solution

\begin{array}{l}
Solution: Direction\ ratios\ of \ the\ lines \ are\ given\ as\ a,b,c\ and\ b-c,\ c-a, a-b.\\
Angle\ between\ the\ lines:\ cos\theta =\ \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}}}\\
a_{1} a_{2} +b_{1} b_{2} +c_{1} c_{2} =a( b-c) +b(c-a)+c(a-b)\\
=ab-ac+bc-ab+ac-ab=0\\
\Longrightarrow cos\theta =0, Hence,\ \theta =90^{0}\\
\end{array}

The NCERT Class 12 Maths Misc. Ex. Ch-11 (3D Geometry) chapter is particularly challenging, requiring extra practice.

Q.2. NCERT Class 12 Maths Misc. Ex. Ch-11 (3D Geometry) Solution

2. Find the equation of a line parallel to the x-axis and passing through the origin.
Solution: The equation of the line parallel to the x-axis and passing through the origin
will\ be:\

Practice problems from the NCERT Class 12 Maths Misc. Ex. Ch-11 (3D Geometry) can greatly enhance your performance.

For students preparing for competitive exams, the NCERT Class 12 Maths Misc. Ex. Ch-11 (3D Geometry) solutions can provide an edge.

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By practicing the NCERT Class 12 Maths Misc. Ex. Ch-11 (3D Geometry) exercises, students can build a solid foundation in geometry.

Q.3. NCERT Class 12 Maths Misc. Ex. Ch-11 (3D Geometry) Solution

Students will find the NCERT Class 12 Maths Misc. Ex. Ch-11 (3D Geometry) solutions invaluable for mastering this chapter.

\begin{array}{l}
3.\ If\ the\ lines\ \frac{x-1}{-3} =\frac{y-2}{2k} =\frac{z-3}{2} \ and\ \frac{x-1}{3k} =\frac{y-1}{1} =\frac{z-6}{5} \ are\ perpendicular\ to\ each\ other.\
Find\ the\ value\ of\ k.
\end{array}

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Understanding the NCERT Class 12 Maths Misc. Ex. Ch-11 (3D Geometry) is essential for students pursuing engineering and science degrees.

Solution for NCERT Class 12 Maths ✨

Miscellaneous Exercise, Ch. 11: 3D-Geometry

Unlock the mysteries of three-dimensional space with our comprehensive solutions for NCERT Class 12 Maths Misc. Ex. Ch-11 (3D Geometry). Dive into the fascinating world of 3D-Geometry, where you’ll explore the intricacies of lines, and angles in space. Our step-by-step solutions make complex problems easy to understand, ensuring you’re well-prepared to tackle any challenge. Embrace the elegance of geometry and master the art of spatial reasoning. These solutions are your key to excelling in exams and beyond! Get free pdf of NCERT Class 12 Maths Misc. Ex. Ch-11 (3D Geometry)

Q.4. NCERT Class 12 Maths Misc. Ex. Ch-11 (3D Geometry) Solution

Make sure to include the NCERT Class 12 Maths Misc. Ex. Ch-11 (3D Geometry) in your daily study schedule.

Q.4. Find the shortest distance between the lines
r = (6,2,2) + λ(1, -2, 2)
and
r = (-4,0,-1) + μ(3, -2, -2).

\begin{array}{l}
Shortest\ Distance\ between\ two\ lines=\ |\frac{\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right) .\left(\overrightarrow{a_{2}} -\overrightarrow{a_{1}}\right)}{|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}} |} |\\
We\ have\ \overrightarrow{b_{1}} =\vec{i} \ -2\vec{j} \ +2\vec{k} \ and\ \overrightarrow{b_{2}} =3\vec{i} -2\vec{j} -2\vec{k}\\
\overrightarrow{b_{1}} \times \overrightarrow{b_{2}} =\ \begin{vmatrix}
\vec{i} & \vec{j} & \vec{k}\\
1 & -2 & 2\\
3 & -2 & -2
\end{vmatrix} =8\vec{i} +8\vec{j} +4\vec{k}\\
|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}} |\ =\ \sqrt{8^{2} +8^{2} +4^{2}} =\sqrt{144} \ =\ 12\\
\overrightarrow{a_{1}} =\ 6\vec{i} +2\vec{j} +2\vec{k} ;\ \overrightarrow{a_{2}} =\ -4\vec{i} -\vec{k}\\
\left(\overrightarrow{a_{2}} -\overrightarrow{a_{1}}\right) =-10\vec{i} -2\vec{j} -3\vec{k}\\
Now,\ |\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right) .\left(\overrightarrow{a_{2}} -\overrightarrow{a_{1}}\right) |=|8\times -10+8\times -2+4\times -3|\\
=|-80-16-12|=|-108|=108\\
Therefore,\ S.\ D.\ =\ \frac{108}{12} =9\ Ans.
\end{array}

Accessing the NCERT Class 12 Maths Misc. Ex. Ch-11 (3D Geometry) solution will aid in clearer conceptual understanding.

Q.5. NCERT Class 12 Maths Misc. Ex. Ch-11 (3D Geometry) Solution

5. Find the vector equation of the line passing through the point (1, 2, – 4) and
   perpendicular to the two lines:

Don’t overlook the NCERT Class 12 Maths Misc. Ex. Ch-11 (3D Geometry) when preparing for your upcoming exams.

\begin{array}{l}
Solution:\
Direction\ vector\ of\ a\ line\ perpendicular\ to\ the\ two\ given\ lines\ will\ be\ cross\ ( vector\ product)\\
of\ the\ direction\ vectors\ of\ the\ two\ lines.\\
Let\ \vec{b} =\overrightarrow{b_{1}} \times \overrightarrow{b_{2}} \ Where\ b_{1} =\ 3\vec{i} -16\vec{j} +7\vec{k} \ and\ \overrightarrow{b_{2}} =3\vec{i} +8\vec{j} -5\vec{k}\\
\ \vec{b} =\begin{vmatrix}
\vec{i} & \vec{j} & \vec{k}\\
3 & -16 & 7\\
3 & 8 & -5
\end{vmatrix} =24\vec{i} +36\vec{j} +72\vec{k} \ \equiv \ 2\vec{i} +3\vec{j} +6\vec{k}\\
Therefore,\ equation\ of\ the\ required\ line\ passing\ through\ ( 1,2,-4)\\
\vec{r} =\vec{i} +2\vec{j} -4\vec{k} \ +\ \lambda ( \ 2\vec{i} +3\vec{j} +6\vec{k})
\end{array}

Conclusion

In conclusion, the NCERT Class 12 Maths Misc. exercise chapter 11, 3D Geometry helps students develop their overall conceptual understanding and analytical thinking skills. This exercise covers important ideas such as direction ratios, direction cosines, angles between lines, equations of lines, and the shortest distance between lines. This chapter not only important for the class 12th board examination but also for the competitive examinations like JEE, ISER, NISER, NDA etc. The questions are asked from this chapter in different competitive examinations.

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