Basics of 3D and 3D Co-ordinate System JEE Main Mathematics: Authentic Shift PYQs (2021-2026) with 2.5-Min Shortcut Tricks & Trap Defense

Correct Answer: c

Explanation: $\frac{3}{5}$ $\begin{matrix}H e r e sin^{2} = 3 sin^{2} \\ B y t h e q u e s t i o n , cos^{2} + cos^{2} + cos^{2} = 1 \\ cos^{2} = 1 2 cos^{2} \\ A d d i n g \left(i\right) a n d \left(i i\right) , w e g e t \\ 1 = 1 + 3 sin^{2} 2 cos^{2} \\ 1 = 1 + 3 \left(1 cos^{2} \right) 2 cos^{2} \\ 5 cos^{2} = 3 \\ cos^{2} = \frac{3}{5}\end{matrix}$

Correct Answer: 8

Explanation: Four diagonals are $O P , \textrm{ } A A ' , \textrm{ } B B ' , \textrm{ } C C "$ , $d c ' s \textrm{ }\text{of}\textrm{ }​ O P \textrm{ }\text{are}\textrm{ } < \textrm{ } \frac{1}{\sqrt{3}} , \textrm{ } \frac{1}{\sqrt{3}} , \textrm{ } \frac{1}{\sqrt{3}} >$  $d c ' s \textrm{ }\text{of}\textrm{ }​ A A ' \textrm{ }\text{are}\textrm{ } < – \textrm{ } \frac{1}{\sqrt{3}} , \textrm{ } \frac{1}{\sqrt{3}} , \textrm{ } \frac{1}{\sqrt{3}} >$  $d c ' s \textrm{ }\text{of}\textrm{ }​ B B ' \textrm{ }\text{are}\textrm{ } < \textrm{ } \frac{1}{\sqrt{3}} , \textrm{ } – \frac{1}{\sqrt{3}} , \textrm{ } \frac{1}{\sqrt{3}} >$  $d c ' s \textrm{ }\text{of}\textrm{ }​ C C ' \textrm{ }\text{are}\textrm{ } < \textrm{ } \frac{1}{\sqrt{3}} , \textrm{ } \frac{1}{\sqrt{3}} , \textrm{ } \frac{– 1}{\sqrt{3}} >$  Let, _l_, _m_, _n_ be the _dc_’_s_ of line which makes the angle a, b, g, d with the four diagonals. $\therefore \textrm{ } cos \textrm{ } \alpha \textrm{ } = \textrm{ } \frac{l \textrm{ } + \textrm{ } m \textrm{ } + \textrm{ } n}{\sqrt{3}} ,$  $cos \textrm{ } \beta \textrm{ } = \textrm{ } \frac{– l \textrm{ } + \textrm{ } m \textrm{ } + \textrm{ } n}{\sqrt{3}} ,$  $cos \textrm{ } \gamma \textrm{ } = \textrm{ } \frac{l \textrm{ } + \textrm{ } m \textrm{ } – \textrm{ } n}{\sqrt{3}} ,$  $cos \textrm{ } \delta \textrm{ } = \textrm{ } \frac{l \textrm{ } + \textrm{ } m \textrm{ } – \textrm{ } n}{\sqrt{3}}$  $\therefore \textrm{ } cos^{2} \alpha \textrm{ } + \textrm{ } cos^{2} \beta \textrm{ } + \textrm{ } cos^{2} \gamma \textrm{ } + \textrm{ } cos^{2} \delta$  $= \textrm{ } \left(\frac{1}{\sqrt{3}}\right)^{2} \textrm{ } \left[1 \textrm{ } + \textrm{ } 1 \textrm{ } + \textrm{ } 1 \textrm{ } + \textrm{ } 1\right] \textrm{ } = \textrm{ } \frac{4}{3}$  $\therefore \textrm{ } sin^{2} \alpha \textrm{ } + \textrm{ } sin^{2} \beta \textrm{ } + \textrm{ } sin^{2} \gamma \textrm{ } + \textrm{ } sin^{2} \delta \textrm{ } = \textrm{ } 4 - \textrm{ } \frac{4}{3} \textrm{ } = \textrm{ } \frac{8}{3}$  Hence, $\lambda = 8$

Correct Answer: N/A

Explanation: $c o s^{ 1} \frac{b^{2}}{\sqrt{\left(a^{2} + b^{2}\right) \left(b^{2} + c^{2}\right)}}$ $\begin{matrix}\begin{matrix}D . R o f A B = \left( a , b , 0\right)\end{matrix} \\ \begin{matrix}D . R o f B C = \left(0 , b , c\right)\end{matrix} \\ \begin{matrix}C o s B = \frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}}\end{matrix} \\ \begin{matrix}= \frac{\left(0 a\right) + \left( b\right) b + 0 c}{\sqrt{a^{2} + b^{2}} \sqrt{b^{2} + c^{2}}}\end{matrix} \\ \begin{matrix}C o s B = \frac{b^{2}}{\sqrt{a^{2} + b^{2}} \sqrt{b^{2} + c^{2}}}\end{matrix} \\ B = c o s^{ 1} \left(\frac{b^{2}}{\sqrt{a^{2} + b^{2}} \sqrt{b^{2} + c^{2}}}\right)\end{matrix}$

Correct Answer: c

Explanation: ​$5 \sqrt{2},\left(\frac{3}{5 \sqrt{2}}, \frac{4}{5 \sqrt{2}}, \frac{1}{\sqrt{2}}\right)$​​ If _a_ is length of line segment then _a_ cos α = 3 ...... (1) a cos β  = 4 .......(2) ​a cos γ  = 5 .......(3)​ Square and add ​$a^{2} \left(\left(cos \right)^{2} \alpha + \left(cos \right)^{2} \beta + \left(cos \right)^{2} \gamma\right) = 50$ $a^{2} \times 1 = 50$ ​$a = 5 \sqrt{2}$ Direction cosines ​$a cos \alpha = 3$​​ ​$cos \alpha = \frac{3}{5 \sqrt{2}}$ ​$a cos \beta = 4$ $cos \beta = \frac{4}{5 \sqrt{2}}$ ​$a cos \gamma = 5$ ​$cos \gamma = \frac{5}{5 \sqrt{2}} = \frac{1}{\sqrt{2}}$

Correct Answer: d

Explanation: ​$1,1, \sqrt{2}$  The intercept form of the plane is given as $\frac{x}{1} + \frac{y}{1} + \frac{z}{c} = 1$​    ​Direction Ratio of normal are $1 , 1 , \frac{1}{c}$ and of given plane are 1, 1, 0. ​$\therefore cos \frac{\pi}{4} = \frac{1.1 + 1.1 + 0. \frac{1}{c}}{\sqrt{1 + 1 + \frac{1}{c^{2}}} \sqrt{1 + 1 + 0}}$ ​$\Rightarrow 2 + \frac{1}{c^{2}} = 4$​ ​​​$\Rightarrow c = \frac{1}{\sqrt{2}}$ $\therefore \&\text{nbsp};\text{DR}’\text{s}\&\text{nbsp};\text{are}\&\text{nbsp}; 1 , 1 , \sqrt{2}$ ​​​