Let S be the set of all real values of lambda | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the four points be $A(-\lambda^2, 1, 1)$,$B(1, -\lambda^2, 1)$,$C(1, 1, -\lambda^2)$,and $D(-1, -1, 1)$.Since the plane passes through all fou

Vedclass · Accepted Answer

$\{\sqrt{3}, -\sqrt{3}\}$

Vedclass · Answer

(B) Let the four points be $A(-\lambda^2, 1, 1)$,$B(1, -\lambda^2, 1)$,$C(1, 1, -\lambda^2)$,and $D(-1, -1, 1)$.Since the plane passes through all four points,they must be coplanar.The condition for four points $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,$(x_3, y_3, z_3)$,and $(x_4, y_4, z_4)$ to be coplanar is given by the determinant:$\begin{vmatrix} x_1-x_4 & y_1-y_4 & z_1-z_4 \ x_2-x_4 & y_2-y_4 & z_2-z_4 \ x_3-x_4 & y_3-y_4 & z_3-z_4 \end{vmatrix} = 0$Substituting the coordinates:$\begin{vmatrix} -\lambda^2+1 & 1+1 & 1-1 \ 1+1 & -\lambda^2+1 & 1-1 \ 1+1 & 1+1 & -\lambda^2-1 \end{vmatrix} = 0$$\begin{vmatrix} 1-\lambda^2 & 2 & 0 \ 2 & 1-\lambda^2 & 0 \ 2 & 2 & -(\lambda^2+1) \end{vmatrix} = 0$Expanding along the third column:$-(\lambda^2+1) \cdot [(1-\lambda^2)^2 - 4] = 0$$-(\lambda^2+1) \cdot (1-\lambda^2-2)(1-\lambda^2+2) = 0$$-(\lambda^2+1) \cdot (-1-\lambda^2)(3-\lambda^2) = 0$$(\lambda^2+1)^2 (3-\lambda^2) = 0$Since $\lambda$ is real,$\lambda^2+1 
eq 0$. Therefore,$3-\lambda^2 = 0$,which gives $\lambda^2 = 3$.Thus,$\lambda = \pm \sqrt{3}$.

Let $S$ be the set of all real values of $\lambda$ such that a plane passing through the points $(-\lambda^2, 1, 1)$,$(1, -\lambda^2, 1)$,and $(1, 1, -\lambda^2)$ also passes through the point $(-1, -1, 1)$. Then $S$ is equal to

Similar Questions

The distance between the planes given by $r \cdot (i + 2j - 2k) + 5 = 0$ and $r \cdot (i + 2j - 2k) - 8 = 0$ is

The plane which bisects the line segment joining the points $A(4, -2, 3)$ and $B(2, 4, -1)$ at right angles also passes through the point:

The plane $ax + by = 0$ is rotated about its line of intersection with the plane $z = 0$ through an angle $\alpha$. Prove that the equation of the plane in its new position is $ax + by \pm (\sqrt{a^{2} + b^{2}} \tan \alpha) z = 0$.

The equation of the plane,passing through the point $(1, 1, 1)$ and perpendicular to the planes $2x + y - 2z = 5$ and $3x - 6y - 2z = 7$,is

The direction ratios of the normal to the plane passing through $(1, 0, 0)$ and $(0, 1, 0)$ which makes an angle $\frac{\pi}{4}$ with the plane $x + y = 3$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module