The sum of all values of $\alpha$, for which the points whose position vectors are $\hat{i} - 2\hat{j} + 3\hat{k}$, $2\hat{i} - 3\hat{j} + 4\hat{k}$, $(\alpha + 1)\hat{i} + 2\hat{k}$ and $9\hat{i} + (\alpha - 8)\hat{j} + 6\hat{k}$ are coplanar, is equal to Options: (A) -2 (B) 2 (C) 6 (D) 4

### Analyzing the Setup To determine the values of $\alpha$ that force the four points $A, B, C,$ and $D$ to lie on a single plane, we must utilize the concept of **coplanarity**. Geometrically, four points are coplanar if the three vectors formed by anchoring them to one point are linearly dependent. We choose point $A$ as our reference. The vectors $\vec{AB}$, $\vec{AC}$, and $\vec{AD}$ must lie within the same plane, which implies that the volume of the parallelepiped they span must be zero. ### The Master Equation The condition for coplanarity is expressed mathematically using the **scalar triple product**: $$[\vec{AB} \ \vec{AC} \ \vec{AD}] = 0$$ Given the coordinates: $A = (1, -2, 3)$ $B = (2, -3, 4)$ $C = (\alpha + 1, 0, 2)$ $D = (9, \alpha - 8, 6)$ We calculate the displacement vectors: $\vec{AB} = (2-1)\hat{i} + (-3 - (-2))\hat{j} + (4-3)\hat{k} = \hat{i} - \hat{j} + \hat{k}$ $\vec{AC} = (\alpha + 1 - 1)\hat{i} + (0 - (-2))\hat{j} + (2-3)\hat{k} = \alpha\hat{i} + 2\hat{j} - \hat{k}$ $\vec{AD} = (9-1)\hat{i} + (\alpha - 8 - (-2))\hat{j} + (6-3)\hat{k} = 8\hat{i} + (\alpha - 6)\hat{j} + 3\hat{k}$ ### The Algebraic Engine To find $\alpha$, we set the determinant of these vectors to zero: $$\begin{vmatrix} 1 & -1 & 1 \\ \alpha & 2 & -1 \\ 8 & \alpha - 6 & 3 \end{vmatrix} = 0$$ Expanding along the first row: $$1[(2)(3) - (-1)(\alpha - 6)] + 1[(\alpha)(3) - (-1)(8)] + 1[(\alpha)(\alpha - 6) - (2)(8)] = 0$$ Simplifying the terms inside the brackets: $$1[6 + \alpha - 6] + 1[3\alpha + 8] + [\alpha^2 - 6\alpha - 16] = 0$$ Combining like terms results in the quadratic equation: $$\alpha^2 - 2\alpha - 8 = 0$$ ### Final Calculation We seek the sum of all possible values of $\alpha$. According to **Vieta's Formulas**, for a quadratic equation of the form $a\alpha^2 + b\alpha + c = 0$, the sum of the roots is given by $-\frac{b}{a}$. In our equation, $a = 1$ and $b = -2$. Therefore: $$\text{Sum of roots} = -\frac{-2}{1} = 2$$ The sum of all values of $\alpha$ that satisfy the condition of coplanarity is **2**.

JEE Main 2023 (06 April Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Vector Algebra: The sum of all values of $α$ , for which the points whose position vectors are $\hat{i} - 2 \hat{j} + 3 \hat{k}$ , $2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ , $(α + 1) \hat{i} + 2 \hat{k}$ and $9 \hat{i} + (α - 8) \hat{j} + 6 \hat{k}$ are coplanar, is equal to

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Imagine four points floating in 3D space. To determine if they lie on a single flat surface (a plane), we need to anchor them. Choose one point as your 'home base' and create three vectors radiating from it. If these three vectors are 'flat'—meaning they don't span any 3D volume—the points are coplanar.

Hint 2

Hint 2 (The Arsenal): Think about the volume of a parallelepiped formed by three vectors

a, b, c

. The volume is given by the scalar triple product

[a b c]

. If the points are coplanar, the 'volume' of the shape they form must be zero. What mathematical tool calculates this volume?

Hint 3

Hint 3 (The Execution): Construct the vectors

A B

A C

, and

A D

by subtracting the position vector of

A

from

B

C

, and

D

respectively. Now, place these vectors as rows in a

3 \times 3

determinant. Set this determinant to zero:

A B_{x} A C_{x} A D_{x} A B_{y} A C_{y} A D_{y} A B_{z} A C_{z} A D_{z} = 0

Visualized Solution

$V i s u a l i z in g t h e P o in t s$

Given position vectors:
$A = \hat{i} - 2 \hat{j} + 3 \hat{k}$
$B = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}$
$C = (α + 1) \hat{i} + 2 \hat{k}$
$D = 9 \hat{i} + (α - 8) \hat{j} + 6 \hat{k}$

$C o n d i t i o n f or C o pl ana r i t y$

Four points $A, B, C, D$ are coplanar if they lie on the same plane.
This implies vectors $A B, A C, A D$ are linearly dependent.
Condition: $[A B A C A D] = 0$

$C a l c u l a t in g A B$

$A B = B - A$
$A B = (2 - 1) \hat{i} + (- 3 - (- 2)) \hat{j} + (4 - 3) \hat{k}$
$A B = \hat{i} - \hat{j} + \hat{k}$

$C a l c u l a t in g A C$

$A C = C - A$
$A C = (α + 1 - 1) \hat{i} + (0 - (- 2)) \hat{j} + (2 - 3) \hat{k}$
$A C = α \hat{i} + 2 \hat{j} - \hat{k}$

$C a l c u l a t in g A D$

$A D = D - A$
$A D = (9 - 1) \hat{i} + (α - 8 - (- 2)) \hat{j} + (6 - 3) \hat{k}$
$A D = 8 \hat{i} + (α - 6) \hat{j} + 3 \hat{k}$

$S e tt in g u p t h e D e t er minan t$

The scalar triple product is the determinant of the components.
$1 α 8 - 1 2 α - 6 1 - 1 3 = 0$

$E x p an d in g t h e D e t er minan t : T er m 1$

Expansion along $R_{1}$ :
Term 1: $1 [(2) (3) - (- 1) (α - 6)]$
$= 1 [6 + α - 6] = α$

$E x p an d in g t h e D e t er minan t : T er m 2$

Term 2: $- (- 1) [(α) (3) - (- 1) (8)]$
$= 1 [3 α + 8] = 3 α + 8$

$E x p an d in g t h e D e t er minan t : T er m 3$

Term 3: $1 [(α) (α - 6) - (2) (8)]$
$= α^{2} - 6 α - 16$

$S im pl i f y in g t o a Q u a d r a t i c E q u a t i o n$

Summing the terms: $α + (3 α + 8) + (α^{2} - 6 α - 16) = 0$
$α^{2} + (1 + 3 - 6) α + (8 - 16) = 0$
$α^{2} - 2 α - 8 = 0$

$F in d in g t h e S u m o f V a l u es$

For $a α^{2} + b α + c = 0$ , sum of roots $= - \frac{b}{a}$
Here $a = 1, b = - 2, c = - 8$
Sum of values $= - \frac{- 2}{1} = 2$

$F ina l C o n c l u s i o n$

Final Answer: The sum of all values of $α$ is 2.
Key Takeaway: Four points are coplanar if the Scalar Triple Product of three vectors formed by them is zero.

00:00 / 00:00

The Sigma Insight: Scalar Triple Product

Solution Diagram

Analyzing the Setup

To determine the values of

α

that force the four points

A, B, C,

and

D

to lie on a single plane, we must utilize the concept of coplanarity. Geometrically, four points are coplanar if the three vectors formed by anchoring them to one point are linearly dependent.

We choose point

A

as our reference. The vectors

A B

A C

, and

A D

must lie within the same plane, which implies that the volume of the parallelepiped they span must be zero.

The Master Equation

The condition for coplanarity is expressed mathematically using the scalar triple product:

[A B A C A D] = 0

Given the coordinates:

A = (1, - 2, 3)

B = (2, - 3, 4)

C = (α + 1, 0, 2)

D = (9, α - 8, 6)

We calculate the displacement vectors:

A B = (2 - 1) \hat{i} + (- 3 - (- 2)) \hat{j} + (4 - 3) \hat{k} = \hat{i} - \hat{j} + \hat{k}

A C = (α + 1 - 1) \hat{i} + (0 - (- 2)) \hat{j} + (2 - 3) \hat{k} = α \hat{i} + 2 \hat{j} - \hat{k}

A D = (9 - 1) \hat{i} + (α - 8 - (- 2)) \hat{j} + (6 - 3) \hat{k} = 8 \hat{i} + (α - 6) \hat{j} + 3 \hat{k}

The Algebraic Engine

To find

α

, we set the determinant of these vectors to zero:

1 α 8 - 1 2 α - 6 1 - 1 3 = 0

Expanding along the first row:

1 [(2) (3) - (- 1) (α - 6)] + 1 [(α) (3) - (- 1) (8)] + 1 [(α) (α - 6) - (2) (8)] = 0

Simplifying the terms inside the brackets:

1 [6 + α - 6] + 1 [3 α + 8] + [α^{2} - 6 α - 16] = 0

Combining like terms results in the quadratic equation:

α^{2} - 2 α - 8 = 0

Final Calculation

We seek the sum of all possible values of

α

. According to Vieta's Formulas, for a quadratic equation of the form

a α^{2} + b α + c = 0

, the sum of the roots is given by

- \frac{b}{a}

In our equation,

a = 1

and

b = - 2

. Therefore:

Sum of roots = - \frac{- 2}{1} = 2

The sum of all values of

α

that satisfy the condition of coplanarity is 2.

Common Traps & FAQs

Wait, does it matter which point I choose as the origin? Could I have used

B

instead of

A

Excellent question! No, it doesn't matter. You could choose

A, B, C,

D

as your reference point. As long as you form three vectors originating from that same point, the scalar triple product will yield the same result. The geometry of the plane remains invariant regardless of your perspective.

Why do we use the scalar triple product? Is there a simpler way?

You could use the equation of a plane, but that's a longer path. The scalar triple product is the 'shortcut' of the vector world. It directly captures the geometric essence of coplanarity—the absence of volume—in one elegant determinant.

I solved the quadratic and got two different values for

α

. Did I do something wrong?

Not at all! That is exactly what the question expects. Sometimes, there are multiple configurations of points that satisfy the condition. You found both, and by summing them, you've captured the full picture.

Select Answer:

Visualized Solution

$V i s u a l i z in g t h e P o in t s$

$C o n d i t i o n f or C o pl ana r i t y$

$C a l c u l a t in g A B$

$C a l c u l a t in g A C$

$C a l c u l a t in g A D$

$S e tt in g u p t h e D e t er minan t$

$E x p an d in g t h e D e t er minan t : T er m 1$

$E x p an d in g t h e D e t er minan t : T er m 2$

$E x p an d in g t h e D e t er minan t : T er m 3$

$S im pl i f y in g t o a Q u a d r a t i c E q u a t i o n$

$F in d in g t h e S u m o f V a l u es$

$F ina l C o n c l u s i o n$

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Master Equation

The Algebraic Engine

Final Calculation

Similar Questions

If the four points, whose position vectors are $3 \hat{i} - 4 \hat{j} + 2 \hat{k}, \hat{i} + 2 \hat{j} - \hat{k}, - 2 \hat{i} - \hat{j} + 3 \hat{k}$ and $5 \hat{i} - 2 α \hat{j} + 4 \hat{k}$ are coplanar, then $α$ is equal to

The sum of the distinct real values of $μ$ , for which the vectors, $μ \hat{i} + \hat{j} + \hat{k}$ , $\hat{i} + μ \hat{j} + \hat{k}$ , $\hat{i} + \hat{j} + μ \hat{k}$ are co-planar, is :

The number of distinct real values of $λ$ , for which the vectors $- λ^{2} \hat{i} + \hat{j} + \hat{k}$ , $\hat{i} - λ^{2} \hat{j} + \hat{k}$ and $\hat{i} + \hat{j} - λ^{2} \hat{k}$ are coplanar, is

If the vectors $a = λ \hat{i} + μ \hat{j} + 4 \hat{k}, b = - 2 \hat{i} + 4 \hat{j} - 2 \hat{k}$ and $c = 2 \hat{i} + 3 \hat{j} + \hat{k}$ are coplanar and the projection of $a$ on the vector $b$ is $54$ units, then the sum of all possible values of $λ + μ$ is equal to

Let $a, b$ and $c$ be three non-zero non-coplanar vectors. Let the position vectors of four points A, B, C and D be $a - b + c$ , $λ a - 3 b + 4 c$ , $- a + 2 b - 3 c$ and $2 a - 4 b + 6 c$ respectively. If AB, AC and AD are coplanar, then $λ$ is :

If four distinct points with position vectors $a, b, c$ and $d$ are coplanar, then $[a b c]$ is equal to

Let a vector $a$ be coplanar with vectors $b = 2 \hat{i} + \hat{j} + \hat{k}$ and $c = \hat{i} - \hat{j} + \hat{k}$ . If $a$ is perpendicular to $d = 3 \hat{i} + 2 \hat{j} + 6 \hat{k}$ , and $∣ a ∣ = 10$ . Then a possible value of $[a b c] + [a b d] + [a c d]$ is equal to:

If the vectors $a \hat{i} + \hat{j} + \hat{k}$ , $\hat{i} + b \hat{j} + \hat{k}$ and $\hat{i} + \hat{j} + c \hat{k}$ ( $a \neq = b \neq = c \neq = 1$ ) are coplanar, then the value of $\frac{1}{( 1 - a )} + \frac{1}{( 1 - b )} + \frac{1}{( 1 - c )} = .........$

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Points

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Coplanarity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating AB

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating AC

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating AD

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Setting up the Determinant

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Expanding the Determinant: Term 1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Expanding the Determinant: Term 2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Expanding the Determinant: Term 3

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying to a Quadratic Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding the Sum of Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Conclusion

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Master Equation

The Algebraic Engine

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the four points, whose position vectors are 3i^−4j^​+2k^,i^+2j^​−k^,−2i^−j^​+3k^ and 5i^−2αj^​+4k^ are coplanar, then α is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of the distinct real values of μ, for which the vectors, μi^+j^​+k^, i^+μj^​+k^, i^+j^​+μk^ are co-planar, is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of distinct real values of λ, for which the vectors −λ2i^+j^​+k^, i^−λ2j^​+k^ and i^+j^​−λ2k^ are coplanar, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If four distinct points with position vectors a,b,c and d are coplanar, then [abc] is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the vectors ai^+j^​+k^, i^+bj^​+k^ and i^+j^​+ck^ (a=b=c=1) are coplanar, then the value of (1−a)1​+(1−b)1​+(1−c)1​=.........

$V i s u a l i z in g t h e P o in t s$

$C o n d i t i o n f or C o pl ana r i t y$

$C a l c u l a t in g A B$

$C a l c u l a t in g A C$

$C a l c u l a t in g A D$

$S e tt in g u p t h e D e t er minan t$

$E x p an d in g t h e D e t er minan t : T er m 1$

$E x p an d in g t h e D e t er minan t : T er m 2$

$E x p an d in g t h e D e t er minan t : T er m 3$

$S im pl i f y in g t o a Q u a d r a t i c E q u a t i o n$

$F in d in g t h e S u m o f V a l u es$

$F ina l C o n c l u s i o n$

If the four points, whose position vectors are $3 \hat{i} - 4 \hat{j} + 2 \hat{k}, \hat{i} + 2 \hat{j} - \hat{k}, - 2 \hat{i} - \hat{j} + 3 \hat{k}$ and $5 \hat{i} - 2 α \hat{j} + 4 \hat{k}$ are coplanar, then $α$ is equal to

The sum of the distinct real values of $μ$ , for which the vectors, $μ \hat{i} + \hat{j} + \hat{k}$ , $\hat{i} + μ \hat{j} + \hat{k}$ , $\hat{i} + \hat{j} + μ \hat{k}$ are co-planar, is :

The number of distinct real values of $λ$ , for which the vectors $- λ^{2} \hat{i} + \hat{j} + \hat{k}$ , $\hat{i} - λ^{2} \hat{j} + \hat{k}$ and $\hat{i} + \hat{j} - λ^{2} \hat{k}$ are coplanar, is

If four distinct points with position vectors $a, b, c$ and $d$ are coplanar, then $[a b c]$ is equal to

If the vectors $a \hat{i} + \hat{j} + \hat{k}$ , $\hat{i} + b \hat{j} + \hat{k}$ and $\hat{i} + \hat{j} + c \hat{k}$ ( $a \neq = b \neq = c \neq = 1$ ) are coplanar, then the value of $\frac{1}{( 1 - a )} + \frac{1}{( 1 - b )} + \frac{1}{( 1 - c )} = .........$