MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Definition of Vector

Let the plane...

Question

Let the plane $a x + b y + c z + d = 0$ bisect the line joining the points $(4, - 3, 1)$ and $(2, 3, - 5)$ at the right angles. If $a, b, c, d$ are integers, then the minimum value of $(a^{2} + b^{2} + c^{2} + d^{2})$ is

Open in App

Solution

Normal of plane $= - - \to P Q = - 2^i + 6^j - 6^k$
$a = - 2, b = 6, c = - 6$
Equation of plane is
$- 2 x + 6 y - 6 z + d = 0$

Mid point of $P (4, - 3, 1)$ and $Q (2, 3, - 5)$ is $M (3, 0, - 2) .$
Since, plane passes through $M,$
$∴ d = - 6$
So, equation of plane is
$- 2 x + 6 y - 6 z - 6 = 0$
$\Rightarrow x - 3 y + 3 z + 3 = 0$
$(a^{2} + b^{2} + c^{2} + d^{2})_{min} = 1^{2} + 9 + 9 + 9 = 28$

Suggest Corrections

thumbs-up

0

Similar questions

Q. If the intercepts made on the axes by the plane which bisects the line joining the points

(1, 2, 3)

(- 3, 4, 5)

at right angles are

(a, 0, 0), (0, b, 0)

(0, 0, c)

(a, b, c)

\frac{x}{a} + \frac{y}{b} = 1

\frac{x}{c} + \frac{y}{d} = 1

intersect the axes at four concyclic points and

a^{2} + c^{2} = b^{2} + d^{2}

, then these lines can intersect at

(a, b, c, d > 0)

Q. If the line segment joining the points

A (a, b)

B (c, d)

subtends an angle

θ

at the origin,then

cos θ = a c + b d \sqrt{(a^{2} + b^{2}) (c^{2} + d^{2})}

a x + b y + c z = d

intersects the coordinate axis at

A, B, C

and the area of the triangle

A B C

(a^{2} + b^{2}) \sqrt{\frac{2 (a^{2} + c^{2} + b^{2})}{a^{2} b^{2} (a^{2} + b^{2})}}

a^{2}, c^{2}, b^{2}

are in A.P. then find the perpendicular distance of the plane from the origin.

a, b, c, d

are the sides of a quadrilateral, then the minimum value of

\frac{a^{2} + b^{2} + c^{2}}{d^{2}}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Introduction

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Definition of Vector

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon