QBManager

Vector Algebra

88194 If $P (3, 2, - 4), Q (9, 8, - 10)$ and $R (5, 4, - 6)$ are collinear, then the ratio in which $R$ divides $P Q$ is

1

1 : 2

2

2 : 1

3

3 : 1

4

1 : 3

Explanation:

(A) : P $(3, 2, - 4), Q (9, 8, - 10)$ and $R (5, 4, - 6)$ are collinear
So, $R$ divides $P Q$ externally and internally.
$(x, y, z) = (\frac{{mx}_{2} + {nx}_{1}}{m + n}, \frac{{my}_{2} + {ny}_{1}}{m + n})$
Therefore,
$Q (x, y, z) = (5, 4, - 6)$
$P (x_{1}, y_{1}, z_{1}) = (3, 2, - 4)$
and $R (x_{1}, y_{2}, z_{2}) = (9, 8, - 10)$
Ratio
$k : 1 = m : n$
So,
$R (5, 4, - 6) = (\frac{k (9) + 3}{k + 1}, \frac{k (8) + 2}{k + 1}, \frac{k (- 10) - 4}{k + 1})$
$(5, 4, - 6) = (\frac{k (9) + 3}{k + 1}, \frac{k (8) + 2}{k + 1}, \frac{k (- 10) - 4}{k + 1})$
So,
$\frac{9 k + 3}{k + 1} = 5 \Rightarrow 9 k + 3 = 5 k + 5$
$9 k - 5 k = 5 - 3 \Rightarrow 4 k = 2$
$k = \frac{2}{4} \Rightarrow k = \frac{1}{2} \Rightarrow$
$∴ k : 1 = 1 : 2$
Hence, required ratio is $1 : 2$

AP EAMCET-2020-23.09.2020

Vector Algebra

88195 The point if intersection of the lines
$l_{1} : r (t) = (I - 6 j + 2 k) + t (i + 2 j + k)$
$l_{2} : R (u) = (4 j + k) + u (2 i + j + 2 k) is$

1

(4, 4, 5)

2

(6, 4, 7)

3

(8, 8, 9)

4

(10, 12, 11)

Explanation:

(C) : Given,
$l_{1} = \vec{r} (t) = (\hat{i} - 6 \hat{j} + 2 \hat{k}) + t (\hat{i} + 2 \hat{j} + \hat{k})$
$l_{2} = \vec{R} (u) = (4 \hat{j} + \hat{k}) + u (2 \hat{i} + j + 2 \hat{k})$
Any point on $l_{1}$ is ${1 + t}, (- 6 + 2 t), (2 + t)$
Any point on $l_{2}$ is $(2 u, 4 + u, 1 + 2 u)$
$∵$ Point of intersection will lies on both
$1 + t = 2 u$
$∴ t = 2 u - 1$
and
$- 6 + 2 t = 4 + u$
$- 6 + 2 (2 u - 1) = 4 + u$
$- 6 + 4 u - 2 - 4 - u = 0$
$3 u = 12$
$u = 12 / 3$
$u = 4$
$Now, t = 2 \times 4 - 1$
$t = 7$
$Hence, point intersection = (1 + 7, - 6 + 14, 2 + 7)$
$= (8, 8, 9)$

AP EAMCET-2012

Vector Algebra

88196 For scalars $λ, μ$ if the vector equation of a plane is
$r = (2 - 3 λ - μ) \hat{i} + (1 - 2 λ + 3 μ) \hat{j} + (- 2 + 2 λ + μ) \hat{k},$
then its Cartesian equation is

1

8 x - 5 y - 7 z + 35 = 0

2

8 x - 5 y + 7 z - 35 = 0

3

8 x + 5 y - 7 z + 35 = 0

4

8 x + 5 y - 7 z - 35 = 0

Explanation:

(D) : Given, vector equation of plane
$r = (2 + 3 λ - μ) \hat{i} + (1 - 2 λ + 3 μ) \hat{j} + (- 2 + 2 λ + μ) k$
Let $r = x \hat{i} + \hat{y} + z \hat{k}$
$∴$ Compare equation. (i) and (ii),
$2 + 3 λ - μ = x$
$1 - 2 λ + 3 μ = y$
$- 2 + 2 λ + μ = z$
Now, equation. $[($ iii) $+ (v)] \Rightarrow 5 λ = x + z$
$λ = \frac{1}{5} (x + z)$
Equation. [(iv) + (v)] $\Rightarrow - 1 + 4 μ = y + z$
$μ = \frac{1}{4} (y + z + 1)$
Putting value of $λ$ and $μ$ in equation (iii)
$2 + \frac{3}{5} (x + z) - \frac{1}{4} (y + z + 1) = x$
$40 + 12 (x + z) - 5 (y + z + 1) = 20 x$
$40 + 12 x + 12 z - 5 y - 5 z - 5 = 20 x$
$- 8 x - 5 y + 7 z + 35 = 0$
$8 x + 5 y - 7 z - 35 = 0$

TS EAMCET-2020-11.09.2020

Vector Algebra

88197 If $b, c$ are non collinear vectors, $| c | \neq 0, a \times (b \times$
c) $+ (a . b) b = (4 - 2 β - \sin α) b + (β^{2} - 1) c$ and (c.c) $a = c$ , then the scalars $α$ and $β$ are

1

α = \frac{π}{2} + \frac{n π}{3}, n \in Z; β = 1

2

α = \frac{π}{2} + 2 n π, n \in Z; β = 1

3

α = \frac{π}{2} + (2 n + 1) \frac{π}{2}, n \in Z, β = 1

4

α = (2 n + 1) \frac{π}{2}, n \in Z, β = \frac{3}{2}

Explanation:

(B) : For non-collinear vectors $b$ and $c | c | \neq 0$ , it is given that, $(c \cdot c) a = c \Rightarrow a \cdot c = 1$
and $a \times (b \times c) + (a \cdot b) b$
$= (4 - 2 β - \sin α) b + (β^{2} - 1) c$
(a.c)b-(a.b)c+(a.b)b $= (4 - 2 β - \sin α)$ $b + (β^{2} - 1) c$
So, $a . c + a . b = 4 - 2 β - \sin α$
and $- a \cdot b = β^{2} - 1$
From Eqs. (i), (ii) and (iii), we get
$\Rightarrow 1 = 4 - 2 β - \sin α + β^{2} - 1$
$(β - 1)^{2} = \sin α = - 1$
So, if $β = 1$ , then $\sin α = 1 \Rightarrow α = 2 n π + \frac{π}{2}, n \in Z$ and other options does not satisfy.

TS EAMCET-2020-11.09.2020

Vector Algebra

88194 If $P (3, 2, - 4), Q (9, 8, - 10)$ and $R (5, 4, - 6)$ are collinear, then the ratio in which $R$ divides $P Q$ is

1

1 : 2

2

2 : 1

3

3 : 1

4

1 : 3

Explanation:

(A) : P $(3, 2, - 4), Q (9, 8, - 10)$ and $R (5, 4, - 6)$ are collinear
So, $R$ divides $P Q$ externally and internally.
$(x, y, z) = (\frac{{mx}_{2} + {nx}_{1}}{m + n}, \frac{{my}_{2} + {ny}_{1}}{m + n})$
Therefore,
$Q (x, y, z) = (5, 4, - 6)$
$P (x_{1}, y_{1}, z_{1}) = (3, 2, - 4)$
and $R (x_{1}, y_{2}, z_{2}) = (9, 8, - 10)$
Ratio
$k : 1 = m : n$
So,
$R (5, 4, - 6) = (\frac{k (9) + 3}{k + 1}, \frac{k (8) + 2}{k + 1}, \frac{k (- 10) - 4}{k + 1})$
$(5, 4, - 6) = (\frac{k (9) + 3}{k + 1}, \frac{k (8) + 2}{k + 1}, \frac{k (- 10) - 4}{k + 1})$
So,
$\frac{9 k + 3}{k + 1} = 5 \Rightarrow 9 k + 3 = 5 k + 5$
$9 k - 5 k = 5 - 3 \Rightarrow 4 k = 2$
$k = \frac{2}{4} \Rightarrow k = \frac{1}{2} \Rightarrow$
$∴ k : 1 = 1 : 2$
Hence, required ratio is $1 : 2$

AP EAMCET-2020-23.09.2020

Vector Algebra

88195 The point if intersection of the lines
$l_{1} : r (t) = (I - 6 j + 2 k) + t (i + 2 j + k)$
$l_{2} : R (u) = (4 j + k) + u (2 i + j + 2 k) is$

1

(4, 4, 5)

2

(6, 4, 7)

3

(8, 8, 9)

4

(10, 12, 11)

Explanation:

(C) : Given,
$l_{1} = \vec{r} (t) = (\hat{i} - 6 \hat{j} + 2 \hat{k}) + t (\hat{i} + 2 \hat{j} + \hat{k})$
$l_{2} = \vec{R} (u) = (4 \hat{j} + \hat{k}) + u (2 \hat{i} + j + 2 \hat{k})$
Any point on $l_{1}$ is ${1 + t}, (- 6 + 2 t), (2 + t)$
Any point on $l_{2}$ is $(2 u, 4 + u, 1 + 2 u)$
$∵$ Point of intersection will lies on both
$1 + t = 2 u$
$∴ t = 2 u - 1$
and
$- 6 + 2 t = 4 + u$
$- 6 + 2 (2 u - 1) = 4 + u$
$- 6 + 4 u - 2 - 4 - u = 0$
$3 u = 12$
$u = 12 / 3$
$u = 4$
$Now, t = 2 \times 4 - 1$
$t = 7$
$Hence, point intersection = (1 + 7, - 6 + 14, 2 + 7)$
$= (8, 8, 9)$

AP EAMCET-2012

Vector Algebra

88196 For scalars $λ, μ$ if the vector equation of a plane is
$r = (2 - 3 λ - μ) \hat{i} + (1 - 2 λ + 3 μ) \hat{j} + (- 2 + 2 λ + μ) \hat{k},$
then its Cartesian equation is

1

8 x - 5 y - 7 z + 35 = 0

2

8 x - 5 y + 7 z - 35 = 0

3

8 x + 5 y - 7 z + 35 = 0

4

8 x + 5 y - 7 z - 35 = 0

Explanation:

(D) : Given, vector equation of plane
$r = (2 + 3 λ - μ) \hat{i} + (1 - 2 λ + 3 μ) \hat{j} + (- 2 + 2 λ + μ) k$
Let $r = x \hat{i} + \hat{y} + z \hat{k}$
$∴$ Compare equation. (i) and (ii),
$2 + 3 λ - μ = x$
$1 - 2 λ + 3 μ = y$
$- 2 + 2 λ + μ = z$
Now, equation. $[($ iii) $+ (v)] \Rightarrow 5 λ = x + z$
$λ = \frac{1}{5} (x + z)$
Equation. [(iv) + (v)] $\Rightarrow - 1 + 4 μ = y + z$
$μ = \frac{1}{4} (y + z + 1)$
Putting value of $λ$ and $μ$ in equation (iii)
$2 + \frac{3}{5} (x + z) - \frac{1}{4} (y + z + 1) = x$
$40 + 12 (x + z) - 5 (y + z + 1) = 20 x$
$40 + 12 x + 12 z - 5 y - 5 z - 5 = 20 x$
$- 8 x - 5 y + 7 z + 35 = 0$
$8 x + 5 y - 7 z - 35 = 0$

TS EAMCET-2020-11.09.2020

Vector Algebra

88197 If $b, c$ are non collinear vectors, $| c | \neq 0, a \times (b \times$
c) $+ (a . b) b = (4 - 2 β - \sin α) b + (β^{2} - 1) c$ and (c.c) $a = c$ , then the scalars $α$ and $β$ are

1

α = \frac{π}{2} + \frac{n π}{3}, n \in Z; β = 1

2

α = \frac{π}{2} + 2 n π, n \in Z; β = 1

3

α = \frac{π}{2} + (2 n + 1) \frac{π}{2}, n \in Z, β = 1

4

α = (2 n + 1) \frac{π}{2}, n \in Z, β = \frac{3}{2}

Explanation:

(B) : For non-collinear vectors $b$ and $c | c | \neq 0$ , it is given that, $(c \cdot c) a = c \Rightarrow a \cdot c = 1$
and $a \times (b \times c) + (a \cdot b) b$
$= (4 - 2 β - \sin α) b + (β^{2} - 1) c$
(a.c)b-(a.b)c+(a.b)b $= (4 - 2 β - \sin α)$ $b + (β^{2} - 1) c$
So, $a . c + a . b = 4 - 2 β - \sin α$
and $- a \cdot b = β^{2} - 1$
From Eqs. (i), (ii) and (iii), we get
$\Rightarrow 1 = 4 - 2 β - \sin α + β^{2} - 1$
$(β - 1)^{2} = \sin α = - 1$
So, if $β = 1$ , then $\sin α = 1 \Rightarrow α = 2 n π + \frac{π}{2}, n \in Z$ and other options does not satisfy.

TS EAMCET-2020-11.09.2020

Vector Algebra

88194 If $P (3, 2, - 4), Q (9, 8, - 10)$ and $R (5, 4, - 6)$ are collinear, then the ratio in which $R$ divides $P Q$ is

1

1 : 2

2

2 : 1

3

3 : 1

4

1 : 3

Explanation:

(A) : P $(3, 2, - 4), Q (9, 8, - 10)$ and $R (5, 4, - 6)$ are collinear
So, $R$ divides $P Q$ externally and internally.
$(x, y, z) = (\frac{{mx}_{2} + {nx}_{1}}{m + n}, \frac{{my}_{2} + {ny}_{1}}{m + n})$
Therefore,
$Q (x, y, z) = (5, 4, - 6)$
$P (x_{1}, y_{1}, z_{1}) = (3, 2, - 4)$
and $R (x_{1}, y_{2}, z_{2}) = (9, 8, - 10)$
Ratio
$k : 1 = m : n$
So,
$R (5, 4, - 6) = (\frac{k (9) + 3}{k + 1}, \frac{k (8) + 2}{k + 1}, \frac{k (- 10) - 4}{k + 1})$
$(5, 4, - 6) = (\frac{k (9) + 3}{k + 1}, \frac{k (8) + 2}{k + 1}, \frac{k (- 10) - 4}{k + 1})$
So,
$\frac{9 k + 3}{k + 1} = 5 \Rightarrow 9 k + 3 = 5 k + 5$
$9 k - 5 k = 5 - 3 \Rightarrow 4 k = 2$
$k = \frac{2}{4} \Rightarrow k = \frac{1}{2} \Rightarrow$
$∴ k : 1 = 1 : 2$
Hence, required ratio is $1 : 2$

AP EAMCET-2020-23.09.2020

Vector Algebra

88195 The point if intersection of the lines
$l_{1} : r (t) = (I - 6 j + 2 k) + t (i + 2 j + k)$
$l_{2} : R (u) = (4 j + k) + u (2 i + j + 2 k) is$

1

(4, 4, 5)

2

(6, 4, 7)

3

(8, 8, 9)

4

(10, 12, 11)

Explanation:

(C) : Given,
$l_{1} = \vec{r} (t) = (\hat{i} - 6 \hat{j} + 2 \hat{k}) + t (\hat{i} + 2 \hat{j} + \hat{k})$
$l_{2} = \vec{R} (u) = (4 \hat{j} + \hat{k}) + u (2 \hat{i} + j + 2 \hat{k})$
Any point on $l_{1}$ is ${1 + t}, (- 6 + 2 t), (2 + t)$
Any point on $l_{2}$ is $(2 u, 4 + u, 1 + 2 u)$
$∵$ Point of intersection will lies on both
$1 + t = 2 u$
$∴ t = 2 u - 1$
and
$- 6 + 2 t = 4 + u$
$- 6 + 2 (2 u - 1) = 4 + u$
$- 6 + 4 u - 2 - 4 - u = 0$
$3 u = 12$
$u = 12 / 3$
$u = 4$
$Now, t = 2 \times 4 - 1$
$t = 7$
$Hence, point intersection = (1 + 7, - 6 + 14, 2 + 7)$
$= (8, 8, 9)$

AP EAMCET-2012

Vector Algebra

88196 For scalars $λ, μ$ if the vector equation of a plane is
$r = (2 - 3 λ - μ) \hat{i} + (1 - 2 λ + 3 μ) \hat{j} + (- 2 + 2 λ + μ) \hat{k},$
then its Cartesian equation is

1

8 x - 5 y - 7 z + 35 = 0

2

8 x - 5 y + 7 z - 35 = 0

3

8 x + 5 y - 7 z + 35 = 0

4

8 x + 5 y - 7 z - 35 = 0

Explanation:

(D) : Given, vector equation of plane
$r = (2 + 3 λ - μ) \hat{i} + (1 - 2 λ + 3 μ) \hat{j} + (- 2 + 2 λ + μ) k$
Let $r = x \hat{i} + \hat{y} + z \hat{k}$
$∴$ Compare equation. (i) and (ii),
$2 + 3 λ - μ = x$
$1 - 2 λ + 3 μ = y$
$- 2 + 2 λ + μ = z$
Now, equation. $[($ iii) $+ (v)] \Rightarrow 5 λ = x + z$
$λ = \frac{1}{5} (x + z)$
Equation. [(iv) + (v)] $\Rightarrow - 1 + 4 μ = y + z$
$μ = \frac{1}{4} (y + z + 1)$
Putting value of $λ$ and $μ$ in equation (iii)
$2 + \frac{3}{5} (x + z) - \frac{1}{4} (y + z + 1) = x$
$40 + 12 (x + z) - 5 (y + z + 1) = 20 x$
$40 + 12 x + 12 z - 5 y - 5 z - 5 = 20 x$
$- 8 x - 5 y + 7 z + 35 = 0$
$8 x + 5 y - 7 z - 35 = 0$

TS EAMCET-2020-11.09.2020

Vector Algebra

88197 If $b, c$ are non collinear vectors, $| c | \neq 0, a \times (b \times$
c) $+ (a . b) b = (4 - 2 β - \sin α) b + (β^{2} - 1) c$ and (c.c) $a = c$ , then the scalars $α$ and $β$ are

1

α = \frac{π}{2} + \frac{n π}{3}, n \in Z; β = 1

2

α = \frac{π}{2} + 2 n π, n \in Z; β = 1

3

α = \frac{π}{2} + (2 n + 1) \frac{π}{2}, n \in Z, β = 1

4

α = (2 n + 1) \frac{π}{2}, n \in Z, β = \frac{3}{2}

Explanation:

(B) : For non-collinear vectors $b$ and $c | c | \neq 0$ , it is given that, $(c \cdot c) a = c \Rightarrow a \cdot c = 1$
and $a \times (b \times c) + (a \cdot b) b$
$= (4 - 2 β - \sin α) b + (β^{2} - 1) c$
(a.c)b-(a.b)c+(a.b)b $= (4 - 2 β - \sin α)$ $b + (β^{2} - 1) c$
So, $a . c + a . b = 4 - 2 β - \sin α$
and $- a \cdot b = β^{2} - 1$
From Eqs. (i), (ii) and (iii), we get
$\Rightarrow 1 = 4 - 2 β - \sin α + β^{2} - 1$
$(β - 1)^{2} = \sin α = - 1$
So, if $β = 1$ , then $\sin α = 1 \Rightarrow α = 2 n π + \frac{π}{2}, n \in Z$ and other options does not satisfy.

TS EAMCET-2020-11.09.2020

Vector Algebra

88194 If $P (3, 2, - 4), Q (9, 8, - 10)$ and $R (5, 4, - 6)$ are collinear, then the ratio in which $R$ divides $P Q$ is

1

1 : 2

2

2 : 1

3

3 : 1

4

1 : 3

Explanation:

(A) : P $(3, 2, - 4), Q (9, 8, - 10)$ and $R (5, 4, - 6)$ are collinear
So, $R$ divides $P Q$ externally and internally.
$(x, y, z) = (\frac{{mx}_{2} + {nx}_{1}}{m + n}, \frac{{my}_{2} + {ny}_{1}}{m + n})$
Therefore,
$Q (x, y, z) = (5, 4, - 6)$
$P (x_{1}, y_{1}, z_{1}) = (3, 2, - 4)$
and $R (x_{1}, y_{2}, z_{2}) = (9, 8, - 10)$
Ratio
$k : 1 = m : n$
So,
$R (5, 4, - 6) = (\frac{k (9) + 3}{k + 1}, \frac{k (8) + 2}{k + 1}, \frac{k (- 10) - 4}{k + 1})$
$(5, 4, - 6) = (\frac{k (9) + 3}{k + 1}, \frac{k (8) + 2}{k + 1}, \frac{k (- 10) - 4}{k + 1})$
So,
$\frac{9 k + 3}{k + 1} = 5 \Rightarrow 9 k + 3 = 5 k + 5$
$9 k - 5 k = 5 - 3 \Rightarrow 4 k = 2$
$k = \frac{2}{4} \Rightarrow k = \frac{1}{2} \Rightarrow$
$∴ k : 1 = 1 : 2$
Hence, required ratio is $1 : 2$

AP EAMCET-2020-23.09.2020

Vector Algebra

88195 The point if intersection of the lines
$l_{1} : r (t) = (I - 6 j + 2 k) + t (i + 2 j + k)$
$l_{2} : R (u) = (4 j + k) + u (2 i + j + 2 k) is$

1

(4, 4, 5)

2

(6, 4, 7)

3

(8, 8, 9)

4

(10, 12, 11)

Explanation:

(C) : Given,
$l_{1} = \vec{r} (t) = (\hat{i} - 6 \hat{j} + 2 \hat{k}) + t (\hat{i} + 2 \hat{j} + \hat{k})$
$l_{2} = \vec{R} (u) = (4 \hat{j} + \hat{k}) + u (2 \hat{i} + j + 2 \hat{k})$
Any point on $l_{1}$ is ${1 + t}, (- 6 + 2 t), (2 + t)$
Any point on $l_{2}$ is $(2 u, 4 + u, 1 + 2 u)$
$∵$ Point of intersection will lies on both
$1 + t = 2 u$
$∴ t = 2 u - 1$
and
$- 6 + 2 t = 4 + u$
$- 6 + 2 (2 u - 1) = 4 + u$
$- 6 + 4 u - 2 - 4 - u = 0$
$3 u = 12$
$u = 12 / 3$
$u = 4$
$Now, t = 2 \times 4 - 1$
$t = 7$
$Hence, point intersection = (1 + 7, - 6 + 14, 2 + 7)$
$= (8, 8, 9)$

AP EAMCET-2012

Vector Algebra

88196 For scalars $λ, μ$ if the vector equation of a plane is
$r = (2 - 3 λ - μ) \hat{i} + (1 - 2 λ + 3 μ) \hat{j} + (- 2 + 2 λ + μ) \hat{k},$
then its Cartesian equation is

1

8 x - 5 y - 7 z + 35 = 0

2

8 x - 5 y + 7 z - 35 = 0

3

8 x + 5 y - 7 z + 35 = 0

4

8 x + 5 y - 7 z - 35 = 0

Explanation:

(D) : Given, vector equation of plane
$r = (2 + 3 λ - μ) \hat{i} + (1 - 2 λ + 3 μ) \hat{j} + (- 2 + 2 λ + μ) k$
Let $r = x \hat{i} + \hat{y} + z \hat{k}$
$∴$ Compare equation. (i) and (ii),
$2 + 3 λ - μ = x$
$1 - 2 λ + 3 μ = y$
$- 2 + 2 λ + μ = z$
Now, equation. $[($ iii) $+ (v)] \Rightarrow 5 λ = x + z$
$λ = \frac{1}{5} (x + z)$
Equation. [(iv) + (v)] $\Rightarrow - 1 + 4 μ = y + z$
$μ = \frac{1}{4} (y + z + 1)$
Putting value of $λ$ and $μ$ in equation (iii)
$2 + \frac{3}{5} (x + z) - \frac{1}{4} (y + z + 1) = x$
$40 + 12 (x + z) - 5 (y + z + 1) = 20 x$
$40 + 12 x + 12 z - 5 y - 5 z - 5 = 20 x$
$- 8 x - 5 y + 7 z + 35 = 0$
$8 x + 5 y - 7 z - 35 = 0$

TS EAMCET-2020-11.09.2020

Vector Algebra

88197 If $b, c$ are non collinear vectors, $| c | \neq 0, a \times (b \times$
c) $+ (a . b) b = (4 - 2 β - \sin α) b + (β^{2} - 1) c$ and (c.c) $a = c$ , then the scalars $α$ and $β$ are

1

α = \frac{π}{2} + \frac{n π}{3}, n \in Z; β = 1

2

α = \frac{π}{2} + 2 n π, n \in Z; β = 1

3

α = \frac{π}{2} + (2 n + 1) \frac{π}{2}, n \in Z, β = 1

4

α = (2 n + 1) \frac{π}{2}, n \in Z, β = \frac{3}{2}

Explanation:

(B) : For non-collinear vectors $b$ and $c | c | \neq 0$ , it is given that, $(c \cdot c) a = c \Rightarrow a \cdot c = 1$
and $a \times (b \times c) + (a \cdot b) b$
$= (4 - 2 β - \sin α) b + (β^{2} - 1) c$
(a.c)b-(a.b)c+(a.b)b $= (4 - 2 β - \sin α)$ $b + (β^{2} - 1) c$
So, $a . c + a . b = 4 - 2 β - \sin α$
and $- a \cdot b = β^{2} - 1$
From Eqs. (i), (ii) and (iii), we get
$\Rightarrow 1 = 4 - 2 β - \sin α + β^{2} - 1$
$(β - 1)^{2} = \sin α = - 1$
So, if $β = 1$ , then $\sin α = 1 \Rightarrow α = 2 n π + \frac{π}{2}, n \in Z$ and other options does not satisfy.

TS EAMCET-2020-11.09.2020

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: