QBManager

Vector Algebra

88185 If $a, b, c$ are non-zero, non-collinear vectors and $\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a}$ , then $\vec{a} + \vec{b} + \vec{c} =$

1

3 \vec{a}

2 0

3

3 (\vec{a} \times \vec{b})

4

3 (\vec{b} \times \vec{c})

Explanation:

(B) : Given, a, b, c are non-zero, non-collinear vectors and $\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a}$
Let,
$\vec{a} + \vec{b} + \vec{c} = k$
$\vec{a} \times (\vec{a} + \vec{b} + \vec{c}) = k \times \vec{a}$
$\vec{a} \times \vec{a} + \vec{a} \times \vec{b} + \vec{a} \times \vec{c} = k \times \vec{a}$
$0 + \vec{a} \times \vec{b} - \vec{c} \times \vec{a} = k \times \vec{a}$
$\vec{a} \times \vec{b} - \vec{a} \times \vec{b} = k \times \vec{a}$
$0 = k \times \vec{a}$
$k = 0$
$∴ \vec{a} + \vec{b} + \vec{c} = 0$

AP EAMCET-2019-21.04.2019

Vector Algebra

88186 If the vectors $p \hat{i} + \hat{j} + \hat{k}, \hat{i} + q \hat{j} + \hat{k}$ and $\hat{i} + \hat{j} + r \hat{k}$ (where, $p \neq q \neq r \neq 1$ ) are coplanar, then the value of $p q r - (p + q + r)$ is

1 -2

2 2

3 0

4 -1

Explanation:

(A) : Given,
$a, b, c$ are coplanar and $p \neq q \neq r \neq 1$
$a = p \hat{i} + \hat{j} + \hat{k}$
$b = \hat{i} + q \hat{j} + \hat{k}$
$c = \hat{i} + \hat{j} + r \hat{k}$
Since, $a, b$ and $c$ are coplanar
$∴ \begin{array}{lll} [a & b & c] \end{array} = 0$
$| \begin{array}{lll} p & 1 & 1 \\ 1 & q & 1 \\ 1 & 1 & r \end{array} | = 0$
$p (qr - 1) - 1 (r - 1) + 1 (1 - q) = 0$
$pqr - p - r + 1 + 1 - q = 0$
$pqr - (p + q + r) + 2 = 0$
$pqr - (p + q + r) = - 2$

JEE Main-2016

Vector Algebra

88187 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 coplanar, is

1 2

2 0

3 1

4 -1

Explanation:

(D) : Given,
Three coplanar vectors
$\vec{a} = μ \hat{i} + \hat{j} + \hat{k}$
$\vec{b} = \hat{i} + μ \hat{j} + \hat{k}$
$\vec{c} = \hat{i} + \hat{j} + μ \hat{k}$
$[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}] = 0$
$| \begin{array}{ccc} μ & 1 & 1 \\ 1 & μ & 1 \\ 1 & 1 & μ \end{array} | = 0$
$μ (μ^{2} - 1) - 1 (μ - 1) + (1 - μ) = 0$
$(μ - 1) {μ (μ + 1) - 1 + (- 1)} = 0$
$(μ - 1) {μ^{2} + μ - 2} = 0$
$(μ - 1) (μ - 1) (μ + 2) = 0$
$μ = 1, 1, - 2$
So, district real values of $μ$ are 1 and -2
Therefore, their sum $= - 1$

JEE Main-2019-12.01.2019

Vector Algebra

88188 Let $a = \hat{i} + 2 \hat{j} + 4 \hat{k}, b = \hat{i} + λ \hat{j} + 4 \hat{k}$ and $c = 2 \hat{i} + 4 \hat{j} + (λ^{2} - 1) \hat{k}$ be coplanar vectors. Then, the non-zero vector $a \times c$ is

1

- 10 \hat{i} + 5 \hat{j}

2

- 10 \hat{i} - 5 \hat{j}

3

- 14 \hat{i} - 5 \hat{j}

4

- 14 \hat{i} + 5 \hat{j}

Explanation:

(A) : Given,
$\vec{a} = \hat{i} + 2 \hat{j} + 4 \hat{k}$
$\vec{b} = \hat{i} + λ \hat{j} + 4 \hat{k}$
$\vec{c} = 2 \hat{i} + 4 \hat{j} + (λ^{2} - 1) \hat{k}$
Since, $\vec{a}, \vec{b}$ and $\vec{c}$ are coplanar,
$∴ | \begin{array}{ccc} 1 & 2 & 4 \\ 1 & λ & 4 \\ 2 & 4 & (λ^{2} - 1) \end{array} | = 0$
$(λ (λ^{2} - 1) - 16) - 2 (λ^{2} - 1 - 8) + 4 (4 - 2 λ) = 0$
$(λ^{3} - λ - 16) - 2 λ^{2} + 18 + 16 - 8 λ = 0$
$λ^{3} - 2 λ^{2} - 9 λ + 18 = 0$
$(λ - 2) (λ - 3) (λ + 3) = 0$
$λ = 2, 3, - 3$
$So, λ = 2 (as \vec{a}, 11 \vec{c} for λ = \pm 3)$
$\vec{a} \times \vec{c} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 4 \\ 2 & 4 & 3 \end{array} |$
$= \hat{i} (6 - 16) - \hat{j} (3 - 8) + \hat{k} (4 - 4) = - 10 \hat{i} + 5 \hat{j}$

JEE Main-2019-11.01.2019

Vector Algebra

88189 Let the vectors $(2 + a + b) \hat{i} + (a + 2 b + c) \hat{j} - (b + c) \hat{k},$
$(1 + b) \hat{i} + 2 b \hat{j} - b \hat{k} and (2 + b) \hat{i} + 2 b \hat{j} + (1 - b) \hat{k},$
$a, b, c \in R$ be coplanar.
Then, which of the following is true?

1

2 b = a + c

2

3 c = a + b

3

a = b + 2 c

4

2 a = b + c

Explanation:

(A) : Given,
$\vec{p} = (2 + a + b) \hat{i} + (a + 2 b + c) \hat{j} - (b + c) \hat{k}$
$\vec{q} = (1 + b) \hat{i} + 2 b \hat{j} - b \hat{k}$
$\vec{r} = (2 + b) \hat{i} + 2 b \hat{j} + (1 - b) \hat{k}$
$\vec{p}, \vec{q} and \vec{r} coplanar,$
$| \begin{array}{ccc} (2 + a + b) & (a + 2 b + c) & - (b + c) \\ (1 + b) & 2 b & - b \\ (2 + b) & 2 b & (1 - b) \end{array} | = 0$
Now,
$R_{3} \to R_{3} - R_{2}, R_{1} \to R_{1} - R_{2}$
So,
$[\begin{array}{ccc} a + 1 & a + c & - c \\ b + 1 & 2 b & - b \\ 1 & 0 & 1 \end{array}] = 0$
$(a + 1) 2 b - (a + c) {b + 1 + b} - c {- 2 b} = 0$
$(a + 1) 2 b - (a + c) (2 b + 1) + 2 bc = 0$
$2 b - a - c = 0$
$2 b = a + c$

JEE Main-2021-25.07.2021

Vector Algebra

88185 If $a, b, c$ are non-zero, non-collinear vectors and $\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a}$ , then $\vec{a} + \vec{b} + \vec{c} =$

1

3 \vec{a}

2 0

3

3 (\vec{a} \times \vec{b})

4

3 (\vec{b} \times \vec{c})

Explanation:

(B) : Given, a, b, c are non-zero, non-collinear vectors and $\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a}$
Let,
$\vec{a} + \vec{b} + \vec{c} = k$
$\vec{a} \times (\vec{a} + \vec{b} + \vec{c}) = k \times \vec{a}$
$\vec{a} \times \vec{a} + \vec{a} \times \vec{b} + \vec{a} \times \vec{c} = k \times \vec{a}$
$0 + \vec{a} \times \vec{b} - \vec{c} \times \vec{a} = k \times \vec{a}$
$\vec{a} \times \vec{b} - \vec{a} \times \vec{b} = k \times \vec{a}$
$0 = k \times \vec{a}$
$k = 0$
$∴ \vec{a} + \vec{b} + \vec{c} = 0$

AP EAMCET-2019-21.04.2019

Vector Algebra

88186 If the vectors $p \hat{i} + \hat{j} + \hat{k}, \hat{i} + q \hat{j} + \hat{k}$ and $\hat{i} + \hat{j} + r \hat{k}$ (where, $p \neq q \neq r \neq 1$ ) are coplanar, then the value of $p q r - (p + q + r)$ is

1 -2

2 2

3 0

4 -1

Explanation:

(A) : Given,
$a, b, c$ are coplanar and $p \neq q \neq r \neq 1$
$a = p \hat{i} + \hat{j} + \hat{k}$
$b = \hat{i} + q \hat{j} + \hat{k}$
$c = \hat{i} + \hat{j} + r \hat{k}$
Since, $a, b$ and $c$ are coplanar
$∴ \begin{array}{lll} [a & b & c] \end{array} = 0$
$| \begin{array}{lll} p & 1 & 1 \\ 1 & q & 1 \\ 1 & 1 & r \end{array} | = 0$
$p (qr - 1) - 1 (r - 1) + 1 (1 - q) = 0$
$pqr - p - r + 1 + 1 - q = 0$
$pqr - (p + q + r) + 2 = 0$
$pqr - (p + q + r) = - 2$

JEE Main-2016

Vector Algebra

88187 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 coplanar, is

1 2

2 0

3 1

4 -1

Explanation:

(D) : Given,
Three coplanar vectors
$\vec{a} = μ \hat{i} + \hat{j} + \hat{k}$
$\vec{b} = \hat{i} + μ \hat{j} + \hat{k}$
$\vec{c} = \hat{i} + \hat{j} + μ \hat{k}$
$[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}] = 0$
$| \begin{array}{ccc} μ & 1 & 1 \\ 1 & μ & 1 \\ 1 & 1 & μ \end{array} | = 0$
$μ (μ^{2} - 1) - 1 (μ - 1) + (1 - μ) = 0$
$(μ - 1) {μ (μ + 1) - 1 + (- 1)} = 0$
$(μ - 1) {μ^{2} + μ - 2} = 0$
$(μ - 1) (μ - 1) (μ + 2) = 0$
$μ = 1, 1, - 2$
So, district real values of $μ$ are 1 and -2
Therefore, their sum $= - 1$

JEE Main-2019-12.01.2019

Vector Algebra

88188 Let $a = \hat{i} + 2 \hat{j} + 4 \hat{k}, b = \hat{i} + λ \hat{j} + 4 \hat{k}$ and $c = 2 \hat{i} + 4 \hat{j} + (λ^{2} - 1) \hat{k}$ be coplanar vectors. Then, the non-zero vector $a \times c$ is

1

- 10 \hat{i} + 5 \hat{j}

2

- 10 \hat{i} - 5 \hat{j}

3

- 14 \hat{i} - 5 \hat{j}

4

- 14 \hat{i} + 5 \hat{j}

Explanation:

(A) : Given,
$\vec{a} = \hat{i} + 2 \hat{j} + 4 \hat{k}$
$\vec{b} = \hat{i} + λ \hat{j} + 4 \hat{k}$
$\vec{c} = 2 \hat{i} + 4 \hat{j} + (λ^{2} - 1) \hat{k}$
Since, $\vec{a}, \vec{b}$ and $\vec{c}$ are coplanar,
$∴ | \begin{array}{ccc} 1 & 2 & 4 \\ 1 & λ & 4 \\ 2 & 4 & (λ^{2} - 1) \end{array} | = 0$
$(λ (λ^{2} - 1) - 16) - 2 (λ^{2} - 1 - 8) + 4 (4 - 2 λ) = 0$
$(λ^{3} - λ - 16) - 2 λ^{2} + 18 + 16 - 8 λ = 0$
$λ^{3} - 2 λ^{2} - 9 λ + 18 = 0$
$(λ - 2) (λ - 3) (λ + 3) = 0$
$λ = 2, 3, - 3$
$So, λ = 2 (as \vec{a}, 11 \vec{c} for λ = \pm 3)$
$\vec{a} \times \vec{c} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 4 \\ 2 & 4 & 3 \end{array} |$
$= \hat{i} (6 - 16) - \hat{j} (3 - 8) + \hat{k} (4 - 4) = - 10 \hat{i} + 5 \hat{j}$

JEE Main-2019-11.01.2019

Vector Algebra

88189 Let the vectors $(2 + a + b) \hat{i} + (a + 2 b + c) \hat{j} - (b + c) \hat{k},$
$(1 + b) \hat{i} + 2 b \hat{j} - b \hat{k} and (2 + b) \hat{i} + 2 b \hat{j} + (1 - b) \hat{k},$
$a, b, c \in R$ be coplanar.
Then, which of the following is true?

1

2 b = a + c

2

3 c = a + b

3

a = b + 2 c

4

2 a = b + c

Explanation:

(A) : Given,
$\vec{p} = (2 + a + b) \hat{i} + (a + 2 b + c) \hat{j} - (b + c) \hat{k}$
$\vec{q} = (1 + b) \hat{i} + 2 b \hat{j} - b \hat{k}$
$\vec{r} = (2 + b) \hat{i} + 2 b \hat{j} + (1 - b) \hat{k}$
$\vec{p}, \vec{q} and \vec{r} coplanar,$
$| \begin{array}{ccc} (2 + a + b) & (a + 2 b + c) & - (b + c) \\ (1 + b) & 2 b & - b \\ (2 + b) & 2 b & (1 - b) \end{array} | = 0$
Now,
$R_{3} \to R_{3} - R_{2}, R_{1} \to R_{1} - R_{2}$
So,
$[\begin{array}{ccc} a + 1 & a + c & - c \\ b + 1 & 2 b & - b \\ 1 & 0 & 1 \end{array}] = 0$
$(a + 1) 2 b - (a + c) {b + 1 + b} - c {- 2 b} = 0$
$(a + 1) 2 b - (a + c) (2 b + 1) + 2 bc = 0$
$2 b - a - c = 0$
$2 b = a + c$

JEE Main-2021-25.07.2021

Vector Algebra

88185 If $a, b, c$ are non-zero, non-collinear vectors and $\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a}$ , then $\vec{a} + \vec{b} + \vec{c} =$

1

3 \vec{a}

2 0

3

3 (\vec{a} \times \vec{b})

4

3 (\vec{b} \times \vec{c})

Explanation:

(B) : Given, a, b, c are non-zero, non-collinear vectors and $\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a}$
Let,
$\vec{a} + \vec{b} + \vec{c} = k$
$\vec{a} \times (\vec{a} + \vec{b} + \vec{c}) = k \times \vec{a}$
$\vec{a} \times \vec{a} + \vec{a} \times \vec{b} + \vec{a} \times \vec{c} = k \times \vec{a}$
$0 + \vec{a} \times \vec{b} - \vec{c} \times \vec{a} = k \times \vec{a}$
$\vec{a} \times \vec{b} - \vec{a} \times \vec{b} = k \times \vec{a}$
$0 = k \times \vec{a}$
$k = 0$
$∴ \vec{a} + \vec{b} + \vec{c} = 0$

AP EAMCET-2019-21.04.2019

Vector Algebra

88186 If the vectors $p \hat{i} + \hat{j} + \hat{k}, \hat{i} + q \hat{j} + \hat{k}$ and $\hat{i} + \hat{j} + r \hat{k}$ (where, $p \neq q \neq r \neq 1$ ) are coplanar, then the value of $p q r - (p + q + r)$ is

1 -2

2 2

3 0

4 -1

Explanation:

(A) : Given,
$a, b, c$ are coplanar and $p \neq q \neq r \neq 1$
$a = p \hat{i} + \hat{j} + \hat{k}$
$b = \hat{i} + q \hat{j} + \hat{k}$
$c = \hat{i} + \hat{j} + r \hat{k}$
Since, $a, b$ and $c$ are coplanar
$∴ \begin{array}{lll} [a & b & c] \end{array} = 0$
$| \begin{array}{lll} p & 1 & 1 \\ 1 & q & 1 \\ 1 & 1 & r \end{array} | = 0$
$p (qr - 1) - 1 (r - 1) + 1 (1 - q) = 0$
$pqr - p - r + 1 + 1 - q = 0$
$pqr - (p + q + r) + 2 = 0$
$pqr - (p + q + r) = - 2$

JEE Main-2016

Vector Algebra

88187 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 coplanar, is

1 2

2 0

3 1

4 -1

Explanation:

(D) : Given,
Three coplanar vectors
$\vec{a} = μ \hat{i} + \hat{j} + \hat{k}$
$\vec{b} = \hat{i} + μ \hat{j} + \hat{k}$
$\vec{c} = \hat{i} + \hat{j} + μ \hat{k}$
$[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}] = 0$
$| \begin{array}{ccc} μ & 1 & 1 \\ 1 & μ & 1 \\ 1 & 1 & μ \end{array} | = 0$
$μ (μ^{2} - 1) - 1 (μ - 1) + (1 - μ) = 0$
$(μ - 1) {μ (μ + 1) - 1 + (- 1)} = 0$
$(μ - 1) {μ^{2} + μ - 2} = 0$
$(μ - 1) (μ - 1) (μ + 2) = 0$
$μ = 1, 1, - 2$
So, district real values of $μ$ are 1 and -2
Therefore, their sum $= - 1$

JEE Main-2019-12.01.2019

Vector Algebra

88188 Let $a = \hat{i} + 2 \hat{j} + 4 \hat{k}, b = \hat{i} + λ \hat{j} + 4 \hat{k}$ and $c = 2 \hat{i} + 4 \hat{j} + (λ^{2} - 1) \hat{k}$ be coplanar vectors. Then, the non-zero vector $a \times c$ is

1

- 10 \hat{i} + 5 \hat{j}

2

- 10 \hat{i} - 5 \hat{j}

3

- 14 \hat{i} - 5 \hat{j}

4

- 14 \hat{i} + 5 \hat{j}

Explanation:

(A) : Given,
$\vec{a} = \hat{i} + 2 \hat{j} + 4 \hat{k}$
$\vec{b} = \hat{i} + λ \hat{j} + 4 \hat{k}$
$\vec{c} = 2 \hat{i} + 4 \hat{j} + (λ^{2} - 1) \hat{k}$
Since, $\vec{a}, \vec{b}$ and $\vec{c}$ are coplanar,
$∴ | \begin{array}{ccc} 1 & 2 & 4 \\ 1 & λ & 4 \\ 2 & 4 & (λ^{2} - 1) \end{array} | = 0$
$(λ (λ^{2} - 1) - 16) - 2 (λ^{2} - 1 - 8) + 4 (4 - 2 λ) = 0$
$(λ^{3} - λ - 16) - 2 λ^{2} + 18 + 16 - 8 λ = 0$
$λ^{3} - 2 λ^{2} - 9 λ + 18 = 0$
$(λ - 2) (λ - 3) (λ + 3) = 0$
$λ = 2, 3, - 3$
$So, λ = 2 (as \vec{a}, 11 \vec{c} for λ = \pm 3)$
$\vec{a} \times \vec{c} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 4 \\ 2 & 4 & 3 \end{array} |$
$= \hat{i} (6 - 16) - \hat{j} (3 - 8) + \hat{k} (4 - 4) = - 10 \hat{i} + 5 \hat{j}$

JEE Main-2019-11.01.2019

Vector Algebra

88189 Let the vectors $(2 + a + b) \hat{i} + (a + 2 b + c) \hat{j} - (b + c) \hat{k},$
$(1 + b) \hat{i} + 2 b \hat{j} - b \hat{k} and (2 + b) \hat{i} + 2 b \hat{j} + (1 - b) \hat{k},$
$a, b, c \in R$ be coplanar.
Then, which of the following is true?

1

2 b = a + c

2

3 c = a + b

3

a = b + 2 c

4

2 a = b + c

Explanation:

(A) : Given,
$\vec{p} = (2 + a + b) \hat{i} + (a + 2 b + c) \hat{j} - (b + c) \hat{k}$
$\vec{q} = (1 + b) \hat{i} + 2 b \hat{j} - b \hat{k}$
$\vec{r} = (2 + b) \hat{i} + 2 b \hat{j} + (1 - b) \hat{k}$
$\vec{p}, \vec{q} and \vec{r} coplanar,$
$| \begin{array}{ccc} (2 + a + b) & (a + 2 b + c) & - (b + c) \\ (1 + b) & 2 b & - b \\ (2 + b) & 2 b & (1 - b) \end{array} | = 0$
Now,
$R_{3} \to R_{3} - R_{2}, R_{1} \to R_{1} - R_{2}$
So,
$[\begin{array}{ccc} a + 1 & a + c & - c \\ b + 1 & 2 b & - b \\ 1 & 0 & 1 \end{array}] = 0$
$(a + 1) 2 b - (a + c) {b + 1 + b} - c {- 2 b} = 0$
$(a + 1) 2 b - (a + c) (2 b + 1) + 2 bc = 0$
$2 b - a - c = 0$
$2 b = a + c$

JEE Main-2021-25.07.2021

Vector Algebra

88185 If $a, b, c$ are non-zero, non-collinear vectors and $\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a}$ , then $\vec{a} + \vec{b} + \vec{c} =$

1

3 \vec{a}

2 0

3

3 (\vec{a} \times \vec{b})

4

3 (\vec{b} \times \vec{c})

Explanation:

(B) : Given, a, b, c are non-zero, non-collinear vectors and $\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a}$
Let,
$\vec{a} + \vec{b} + \vec{c} = k$
$\vec{a} \times (\vec{a} + \vec{b} + \vec{c}) = k \times \vec{a}$
$\vec{a} \times \vec{a} + \vec{a} \times \vec{b} + \vec{a} \times \vec{c} = k \times \vec{a}$
$0 + \vec{a} \times \vec{b} - \vec{c} \times \vec{a} = k \times \vec{a}$
$\vec{a} \times \vec{b} - \vec{a} \times \vec{b} = k \times \vec{a}$
$0 = k \times \vec{a}$
$k = 0$
$∴ \vec{a} + \vec{b} + \vec{c} = 0$

AP EAMCET-2019-21.04.2019

Vector Algebra

88186 If the vectors $p \hat{i} + \hat{j} + \hat{k}, \hat{i} + q \hat{j} + \hat{k}$ and $\hat{i} + \hat{j} + r \hat{k}$ (where, $p \neq q \neq r \neq 1$ ) are coplanar, then the value of $p q r - (p + q + r)$ is

1 -2

2 2

3 0

4 -1

Explanation:

(A) : Given,
$a, b, c$ are coplanar and $p \neq q \neq r \neq 1$
$a = p \hat{i} + \hat{j} + \hat{k}$
$b = \hat{i} + q \hat{j} + \hat{k}$
$c = \hat{i} + \hat{j} + r \hat{k}$
Since, $a, b$ and $c$ are coplanar
$∴ \begin{array}{lll} [a & b & c] \end{array} = 0$
$| \begin{array}{lll} p & 1 & 1 \\ 1 & q & 1 \\ 1 & 1 & r \end{array} | = 0$
$p (qr - 1) - 1 (r - 1) + 1 (1 - q) = 0$
$pqr - p - r + 1 + 1 - q = 0$
$pqr - (p + q + r) + 2 = 0$
$pqr - (p + q + r) = - 2$

JEE Main-2016

Vector Algebra

88187 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 coplanar, is

1 2

2 0

3 1

4 -1

Explanation:

(D) : Given,
Three coplanar vectors
$\vec{a} = μ \hat{i} + \hat{j} + \hat{k}$
$\vec{b} = \hat{i} + μ \hat{j} + \hat{k}$
$\vec{c} = \hat{i} + \hat{j} + μ \hat{k}$
$[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}] = 0$
$| \begin{array}{ccc} μ & 1 & 1 \\ 1 & μ & 1 \\ 1 & 1 & μ \end{array} | = 0$
$μ (μ^{2} - 1) - 1 (μ - 1) + (1 - μ) = 0$
$(μ - 1) {μ (μ + 1) - 1 + (- 1)} = 0$
$(μ - 1) {μ^{2} + μ - 2} = 0$
$(μ - 1) (μ - 1) (μ + 2) = 0$
$μ = 1, 1, - 2$
So, district real values of $μ$ are 1 and -2
Therefore, their sum $= - 1$

JEE Main-2019-12.01.2019

Vector Algebra

88188 Let $a = \hat{i} + 2 \hat{j} + 4 \hat{k}, b = \hat{i} + λ \hat{j} + 4 \hat{k}$ and $c = 2 \hat{i} + 4 \hat{j} + (λ^{2} - 1) \hat{k}$ be coplanar vectors. Then, the non-zero vector $a \times c$ is

1

- 10 \hat{i} + 5 \hat{j}

2

- 10 \hat{i} - 5 \hat{j}

3

- 14 \hat{i} - 5 \hat{j}

4

- 14 \hat{i} + 5 \hat{j}

Explanation:

(A) : Given,
$\vec{a} = \hat{i} + 2 \hat{j} + 4 \hat{k}$
$\vec{b} = \hat{i} + λ \hat{j} + 4 \hat{k}$
$\vec{c} = 2 \hat{i} + 4 \hat{j} + (λ^{2} - 1) \hat{k}$
Since, $\vec{a}, \vec{b}$ and $\vec{c}$ are coplanar,
$∴ | \begin{array}{ccc} 1 & 2 & 4 \\ 1 & λ & 4 \\ 2 & 4 & (λ^{2} - 1) \end{array} | = 0$
$(λ (λ^{2} - 1) - 16) - 2 (λ^{2} - 1 - 8) + 4 (4 - 2 λ) = 0$
$(λ^{3} - λ - 16) - 2 λ^{2} + 18 + 16 - 8 λ = 0$
$λ^{3} - 2 λ^{2} - 9 λ + 18 = 0$
$(λ - 2) (λ - 3) (λ + 3) = 0$
$λ = 2, 3, - 3$
$So, λ = 2 (as \vec{a}, 11 \vec{c} for λ = \pm 3)$
$\vec{a} \times \vec{c} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 4 \\ 2 & 4 & 3 \end{array} |$
$= \hat{i} (6 - 16) - \hat{j} (3 - 8) + \hat{k} (4 - 4) = - 10 \hat{i} + 5 \hat{j}$

JEE Main-2019-11.01.2019

Vector Algebra

88189 Let the vectors $(2 + a + b) \hat{i} + (a + 2 b + c) \hat{j} - (b + c) \hat{k},$
$(1 + b) \hat{i} + 2 b \hat{j} - b \hat{k} and (2 + b) \hat{i} + 2 b \hat{j} + (1 - b) \hat{k},$
$a, b, c \in R$ be coplanar.
Then, which of the following is true?

1

2 b = a + c

2

3 c = a + b

3

a = b + 2 c

4

2 a = b + c

Explanation:

(A) : Given,
$\vec{p} = (2 + a + b) \hat{i} + (a + 2 b + c) \hat{j} - (b + c) \hat{k}$
$\vec{q} = (1 + b) \hat{i} + 2 b \hat{j} - b \hat{k}$
$\vec{r} = (2 + b) \hat{i} + 2 b \hat{j} + (1 - b) \hat{k}$
$\vec{p}, \vec{q} and \vec{r} coplanar,$
$| \begin{array}{ccc} (2 + a + b) & (a + 2 b + c) & - (b + c) \\ (1 + b) & 2 b & - b \\ (2 + b) & 2 b & (1 - b) \end{array} | = 0$
Now,
$R_{3} \to R_{3} - R_{2}, R_{1} \to R_{1} - R_{2}$
So,
$[\begin{array}{ccc} a + 1 & a + c & - c \\ b + 1 & 2 b & - b \\ 1 & 0 & 1 \end{array}] = 0$
$(a + 1) 2 b - (a + c) {b + 1 + b} - c {- 2 b} = 0$
$(a + 1) 2 b - (a + c) (2 b + 1) + 2 bc = 0$
$2 b - a - c = 0$
$2 b = a + c$

JEE Main-2021-25.07.2021

Vector Algebra

88185 If $a, b, c$ are non-zero, non-collinear vectors and $\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a}$ , then $\vec{a} + \vec{b} + \vec{c} =$

1

3 \vec{a}

2 0

3

3 (\vec{a} \times \vec{b})

4

3 (\vec{b} \times \vec{c})

Explanation:

(B) : Given, a, b, c are non-zero, non-collinear vectors and $\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a}$
Let,
$\vec{a} + \vec{b} + \vec{c} = k$
$\vec{a} \times (\vec{a} + \vec{b} + \vec{c}) = k \times \vec{a}$
$\vec{a} \times \vec{a} + \vec{a} \times \vec{b} + \vec{a} \times \vec{c} = k \times \vec{a}$
$0 + \vec{a} \times \vec{b} - \vec{c} \times \vec{a} = k \times \vec{a}$
$\vec{a} \times \vec{b} - \vec{a} \times \vec{b} = k \times \vec{a}$
$0 = k \times \vec{a}$
$k = 0$
$∴ \vec{a} + \vec{b} + \vec{c} = 0$

AP EAMCET-2019-21.04.2019

Vector Algebra

88186 If the vectors $p \hat{i} + \hat{j} + \hat{k}, \hat{i} + q \hat{j} + \hat{k}$ and $\hat{i} + \hat{j} + r \hat{k}$ (where, $p \neq q \neq r \neq 1$ ) are coplanar, then the value of $p q r - (p + q + r)$ is

1 -2

2 2

3 0

4 -1

Explanation:

(A) : Given,
$a, b, c$ are coplanar and $p \neq q \neq r \neq 1$
$a = p \hat{i} + \hat{j} + \hat{k}$
$b = \hat{i} + q \hat{j} + \hat{k}$
$c = \hat{i} + \hat{j} + r \hat{k}$
Since, $a, b$ and $c$ are coplanar
$∴ \begin{array}{lll} [a & b & c] \end{array} = 0$
$| \begin{array}{lll} p & 1 & 1 \\ 1 & q & 1 \\ 1 & 1 & r \end{array} | = 0$
$p (qr - 1) - 1 (r - 1) + 1 (1 - q) = 0$
$pqr - p - r + 1 + 1 - q = 0$
$pqr - (p + q + r) + 2 = 0$
$pqr - (p + q + r) = - 2$

JEE Main-2016

Vector Algebra

88187 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 coplanar, is

1 2

2 0

3 1

4 -1

Explanation:

(D) : Given,
Three coplanar vectors
$\vec{a} = μ \hat{i} + \hat{j} + \hat{k}$
$\vec{b} = \hat{i} + μ \hat{j} + \hat{k}$
$\vec{c} = \hat{i} + \hat{j} + μ \hat{k}$
$[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}] = 0$
$| \begin{array}{ccc} μ & 1 & 1 \\ 1 & μ & 1 \\ 1 & 1 & μ \end{array} | = 0$
$μ (μ^{2} - 1) - 1 (μ - 1) + (1 - μ) = 0$
$(μ - 1) {μ (μ + 1) - 1 + (- 1)} = 0$
$(μ - 1) {μ^{2} + μ - 2} = 0$
$(μ - 1) (μ - 1) (μ + 2) = 0$
$μ = 1, 1, - 2$
So, district real values of $μ$ are 1 and -2
Therefore, their sum $= - 1$

JEE Main-2019-12.01.2019

Vector Algebra

88188 Let $a = \hat{i} + 2 \hat{j} + 4 \hat{k}, b = \hat{i} + λ \hat{j} + 4 \hat{k}$ and $c = 2 \hat{i} + 4 \hat{j} + (λ^{2} - 1) \hat{k}$ be coplanar vectors. Then, the non-zero vector $a \times c$ is

1

- 10 \hat{i} + 5 \hat{j}

2

- 10 \hat{i} - 5 \hat{j}

3

- 14 \hat{i} - 5 \hat{j}

4

- 14 \hat{i} + 5 \hat{j}

Explanation:

(A) : Given,
$\vec{a} = \hat{i} + 2 \hat{j} + 4 \hat{k}$
$\vec{b} = \hat{i} + λ \hat{j} + 4 \hat{k}$
$\vec{c} = 2 \hat{i} + 4 \hat{j} + (λ^{2} - 1) \hat{k}$
Since, $\vec{a}, \vec{b}$ and $\vec{c}$ are coplanar,
$∴ | \begin{array}{ccc} 1 & 2 & 4 \\ 1 & λ & 4 \\ 2 & 4 & (λ^{2} - 1) \end{array} | = 0$
$(λ (λ^{2} - 1) - 16) - 2 (λ^{2} - 1 - 8) + 4 (4 - 2 λ) = 0$
$(λ^{3} - λ - 16) - 2 λ^{2} + 18 + 16 - 8 λ = 0$
$λ^{3} - 2 λ^{2} - 9 λ + 18 = 0$
$(λ - 2) (λ - 3) (λ + 3) = 0$
$λ = 2, 3, - 3$
$So, λ = 2 (as \vec{a}, 11 \vec{c} for λ = \pm 3)$
$\vec{a} \times \vec{c} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 4 \\ 2 & 4 & 3 \end{array} |$
$= \hat{i} (6 - 16) - \hat{j} (3 - 8) + \hat{k} (4 - 4) = - 10 \hat{i} + 5 \hat{j}$

JEE Main-2019-11.01.2019

Vector Algebra

88189 Let the vectors $(2 + a + b) \hat{i} + (a + 2 b + c) \hat{j} - (b + c) \hat{k},$
$(1 + b) \hat{i} + 2 b \hat{j} - b \hat{k} and (2 + b) \hat{i} + 2 b \hat{j} + (1 - b) \hat{k},$
$a, b, c \in R$ be coplanar.
Then, which of the following is true?

1

2 b = a + c

2

3 c = a + b

3

a = b + 2 c

4

2 a = b + c

Explanation:

(A) : Given,
$\vec{p} = (2 + a + b) \hat{i} + (a + 2 b + c) \hat{j} - (b + c) \hat{k}$
$\vec{q} = (1 + b) \hat{i} + 2 b \hat{j} - b \hat{k}$
$\vec{r} = (2 + b) \hat{i} + 2 b \hat{j} + (1 - b) \hat{k}$
$\vec{p}, \vec{q} and \vec{r} coplanar,$
$| \begin{array}{ccc} (2 + a + b) & (a + 2 b + c) & - (b + c) \\ (1 + b) & 2 b & - b \\ (2 + b) & 2 b & (1 - b) \end{array} | = 0$
Now,
$R_{3} \to R_{3} - R_{2}, R_{1} \to R_{1} - R_{2}$
So,
$[\begin{array}{ccc} a + 1 & a + c & - c \\ b + 1 & 2 b & - b \\ 1 & 0 & 1 \end{array}] = 0$
$(a + 1) 2 b - (a + c) {b + 1 + b} - c {- 2 b} = 0$
$(a + 1) 2 b - (a + c) (2 b + 1) + 2 bc = 0$
$2 b - a - c = 0$
$2 b = a + c$

JEE Main-2021-25.07.2021