If vectors a, b, c are coplanar, show...

MathematicsScalar Triple ProductJEE Advanced 1989Moderate

If vectors $a, b, c$ are coplanar, show that $a \cdot a b \cdot a c \cdot a a \cdot b b \cdot b c \cdot b a \cdot c b \cdot c c \cdot c = 0$ .

Visualized Solution (Hindi)

Visualizing Coplanar Vectors

Given vectors $a, b, c$ are coplanar.
This means they all lie in the same two-dimensional plane.

The Condition of Linear Dependence

Since $a, b, c$ are coplanar, they are linearly dependent.
There exist scalars $x, y, z$ (not all zero) such that: $x a + y b + z c = 0$

Dot Product with Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Taking the dot product of the equation with $a$ :
$a \cdot (x a + y b + z c) = a \cdot 0$
Expanding: $x (a \cdot a) + y (a \cdot b) + z (a \cdot c) = 0$

Dot Product with Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Taking the dot product of the equation with $b$ :
$b \cdot (x a + y b + z c) = b \cdot 0$
Expanding: $x (b \cdot a) + y (b \cdot b) + z (b \cdot c) = 0$

Dot Product with Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Taking the dot product of the equation with $c$ :
$c \cdot (x a + y b + z c) = c \cdot 0$
Expanding: $x (c \cdot a) + y (c \cdot b) + z (c \cdot c) = 0$

The Homogeneous System

We have a system of three homogeneous equations:
1. $x (a \cdot a) + y (a \cdot b) + z (a \cdot c) = 0$
2. $x (b \cdot a) + y (b \cdot b) + z (b \cdot c) = 0$
3. $x (c \cdot a) + y (c \cdot b) + z (c \cdot c) = 0$

The Determinant Condition

For a non-trivial solution $(x, y, z) \neq = (0, 0, 0)$ , the determinant of the coefficient matrix must be zero:
$a \cdot a b \cdot a c \cdot a a \cdot b b \cdot b c \cdot b a \cdot c b \cdot c c \cdot c = 0$
Hence Proved.

Summary and Beyond

Key Takeaway: The Gram determinant of coplanar vectors is always zero.
Connection: This determinant is equal to $[a b c]^{2}$ .
Next Challenge: What happens to this determinant if the vectors are mutually perpendicular?

00:00 / 00:00

Conceptually Similar Problems

MathematicsVector Triple ProductJEE Advanced 1994Moderate

View in:English Hindi

If the vectors $b, c, d$ are not coplanar, then prove that the vector $(a \times b) \times (c \times d) + (a \times c) \times (d \times b) + (a \times d) \times (b \times c)$ is parallel to $a$ .

MathematicsScalar Triple ProductJEE Advanced 1995SModerate

View in:English Hindi

If $a, b$ and $c$ are three non coplanar vectors, then $(a + b + c) \cdot [(a + b) \times (a + c)]$ equals

(A)

0

(B)

[a b c]

(C)

2 [a b c]

(D)

- [a b c]

Answer:D

MathematicsScalar Triple ProductJEE Advanced 1988Easy

View in:English Hindi

Let $a, b, c$ be three non-coplanar vectors and $p, q, r$ are vectors defined by the relations $p = \frac{b \times c}{[ a b c ]}, q = \frac{c \times a}{[ a b c ]}, r = \frac{a \times b}{[ a b c ]}$ then the value of the expression $(a + b) \cdot p + (b + c) \cdot q + (c + a) \cdot r$ is equal to

(A)

(B)

(C)

(D)

Answer:D

MathematicsVector Triple ProductJEE Advanced 1997Moderate

View in:English Hindi

If $A, B$ and $C$ are vectors such that $∣ B ∣ = ∣ C ∣$ . Prove that $[(A + B) \times (A + C)] \times (B \times C) \cdot (B + C) = 0$ .

MathematicsVector (Cross) ProductJEE Advanced 2004Moderate

View in:English Hindi

If $a, b, c$ and $d$ are distinct vectors such that $a \times c = b \times d$ and $a \times b = c \times d$ . Prove that $(a - d) \cdot (b - c) \neq = 0$ i.e. $a \cdot b + d \cdot c \neq = d \cdot b + a \cdot c$ .

MathematicsScalar Triple ProductJEE Advanced 1985Easy

View in:English Hindi

If $A, B, C$ are three non-coplanar vectors, then $\frac{A \cdot B \times C}{C \times A \cdot B} + \frac{B \cdot A \times C}{C \cdot A \times B} = .........$

Answer:0

MathematicsScalar Triple ProductJEE Advanced 1987Moderate

View in:English Hindi

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 $\frac{1}{( 1 - a )} + \frac{1}{( 1 - b )} + \frac{1}{( 1 - c )} = .........$

Answer:1

MathematicsScalar Triple ProductJEE Main 2003Moderate

View in:English Hindi

If $u, v$ and $w$ are three non-coplanar vectors, then $(u + v - w) \cdot [(u - v) \times (v - w)]$ equals

(A)

3 u \cdot v \times w

(B)

(C)

u \cdot v \times w

(D)

u \cdot w \times v

Answer:C

MathematicsProperties of DeterminantsJEE Main 2003Easy

View in:English Hindi

If $a b c a^{2} b^{2} c^{2} 1 + a^{3} 1 + b^{3} 1 + c^{3} = 0$ and vectors $(1, a, a^{2}), (1, b, b^{2})$ and $(1, c, c^{2})$ are non-coplanar, then the product $ab c$ equals

(A)

(B)

(C)

-1

(D)

Answer:C

MathematicsComponents of a VectorJEE Advanced 1999Moderate

View in:English Hindi

Let $a = 2 \hat{i} + \hat{j} + \hat{k}, b = \hat{i} + 2 \hat{j} - \hat{k}$ and a unit vector $c$ be coplanar. If $c$ is perpendicular to $a$ , then $c =$

(A)

\frac{1}{2} (- \hat{j} + \hat{k})

(B)

\frac{1}{3} (- \hat{i} - \hat{j} - \hat{k})

(C)

\frac{1}{5} (\hat{i} - 2 \hat{j})

(D)

\frac{1}{3} (\hat{i} - \hat{j} - \hat{k})

Answer:A