Let a, b, c be real numbers with a^2 +...

MathematicsProperties of DeterminantsJEE Advanced 2001Moderate

Let $a, b, c$ be real numbers with $a^{2} + b^{2} + c^{2} = 1$ . Show that the equation $a x - b y - c b x + a y c x + a b x + a y - a x + b y - c cy + b c x + a cy + b - a x - b y + c = 0$ represents a straight line.

Visualized Solution (Hindi)

Understanding the Problem

Given: $a, b, c \in R$ and $a^{2} + b^{2} + c^{2} = 1$
Equation: $a x - b y - c b x + a y c x + a b x + a y - a x + b y - c cy + b c x + a cy + b - a x - b y + c = 0$
Objective: Prove the equation represents a straight line.

Strategic Row Operation

Apply row operation: $R_{1} \to a R_{1} + b R_{2} + c R_{3}$
To maintain equality, divide the determinant by $a$ : $Δ = \frac{1}{a} Δ^{'}$

Calculating the First Element

New $R_{11} = a (a x - b y - c) + b (b x + a y) + c (c x + a)$
Expanding: $a^{2} x - ab y - a c + b^{2} x + ab y + c^{2} x + a c$
Simplifying: $x (a^{2} + b^{2} + c^{2}) = x (1) = x$

Simplifying Row 1

Similarly, $R_{12} = y (a^{2} + b^{2} + c^{2}) = y$
And $R_{13} = 1 (a^{2} + b^{2} + c^{2}) = 1$
Resulting Determinant: $\frac{1}{a} x b x + a y c x + a y - a x + b y - c cy + b 1 cy + b - a x - b y + c = 0$

Cleaning Up Rows 2 and 3

Apply operations: $R_{2} \to R_{2} - b R_{1}$ and $R_{3} \to R_{3} - c R_{1}$
This eliminates $b x$ from $R_{2}$ and $c x$ from $R_{3}$ .

The Simplified Determinant

Simplified Determinant: $\frac{1}{a} x a y a y - a x - c b 1 cy - a x - b y = 0$

Expansion Phase

Expand along $R_{1}$ :
$x [(- a x - c) (- a x - b y) - b cy] - y [a y (- a x - b y) - a cy] + 1 [ab y - a (- a x - c)]$

Algebraic Simplification

Expanding terms: $a^{2} x^{3} + ab x^{2} y + a c x^{2} + a^{2} x y^{2} + ab y^{3} + a c y^{2} + ab y + a^{2} x + a c$
Factoring: $a (x^{2} + y^{2} + 1) (a x + b y + c)$

The Final Equation

Final simplified equation: $(x^{2} + y^{2} + 1) (a x + b y + c) = 0$
Since $x, y \in R$ , $x^{2} + y^{2} + 1 \geq 1$
Therefore, $x^{2} + y^{2} + 1 \neq = 0$

Conclusion: A Straight Line

Conclusion: $a x + b y + c = 0$
This is a linear equation in $x$ and $y$ .
Hence, the equation represents a straight line.

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Let $a, b, c$ be any real numbers. Suppose that there are real numbers $x, y, z$ not all zero such that $x = cy + b z, y = a z + c x$ , and $z = b x + a y$ . Then $a^{2} + b^{2} + c^{2} + 2 ab c$ is equal to

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Comprehension Passage

Let

a, b

and

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be three real numbers satisfying

[a b c] 187923777 = [000] \dots (E)

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If the point $P (a, b, c)$ , with reference to (E), lies on the plane $2 x + y + z = 1$ , then the value of $7 a + b + c$ is

(A)

(B)

(C)

(D)

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Let $ω$ be a solution of $x^{3} - 1 = 0$ with $Im (ω) > 0$ , if $a = 2$ with $b$ and $c$ satisfying (E), then the value of $\frac{3}{ω ^{a}} + \frac{1}{ω ^{b}} + \frac{3}{ω ^{c}}$ is equal to

(A)

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(B)

(C)

(D)

-3

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Let $b = 6$ , with $a$ and $c$ satisfying (E). If $α$ and $β$ are the roots of the quadratic equation $a x^{2} + b x + c = 0$ , then $\sum_{n = 0}^{\infty} (\frac{1}{α} + \frac{1}{β})^{n}$ is

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Show that $^{x} C_{r}^{y} C_{r}^{z} C_{r}^{x} C_{r + 1}^{y} C_{r + 1}^{z} C_{r + 1}^{x} C_{r + 2}^{y} C_{r + 2}^{z} C_{r + 2} =^{x} C_{r}^{y} C_{r}^{z} C_{r}^{x + 1} C_{r + 1}^{y + 1} C_{r + 1}^{z + 1} C_{r + 1}^{x + 2} C_{r + 2}^{y + 2} C_{r + 2}^{z + 2} C_{r + 2}$ .

Answer:Proof using

^{n} C_{r} +^{n} C_{r + 1} =^{n + 1} C_{r + 1}

$a x - b y - c$	$b x + a y$	$c x + a$
$b x + a y$	$- a x + b y - c$	$cy + b$
$c x + a$	$cy + b$	$- a x - b y + c$

$x$	$y$	$1$
$b x + a y$	$- a x + b y - c$	$cy + b$
$c x + a$	$cy + b$	$- a x - b y + c$

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a,b,c be real numbers with a2+b2+c2=1. Show that the equation ​ax−by−cbx+aycx+a​bx+ay−ax+by−ccy+b​cx+acy+b−ax−by+c​​=0 represents a straight line.

Visualized Solution (Hindi)

Understanding the Problem

Strategic Row Operation

Calculating the First Element

Simplifying Row 1

Cleaning Up Rows 2 and 3

The Simplified Determinant

Expansion Phase

Algebraic Simplification

The Final Equation

Conclusion: A Straight Line

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