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Vector Algebra

88198 $l, m, n$ are three unit vectors in a right handed system and $L$ is a line through the points $A, B$ , C whose position vectors are $p l + 7 m - 6 n, 2 l +$ $5 m - 4 n$ and $l + 4 m - 3 n$ respectively. If the equation of the plane containing $L$ and the points $(- p, p, p + 1)$ is $a x + by + cz = l$ , then $p (a$ $+ b + c) =$

1 0

2

\frac{- 40}{19}

3

\frac{40}{19}

4 -6

Explanation:

(B) : Let $l, m, n$ be $\hat{i}, \hat{j}, \hat{k}$ $A (p \hat{i} + 7 \hat{j} - 6 \hat{k}), B (2 \hat{i} + 5 \hat{j} - 4 \hat{k})$ and $C (\hat{i} + 4 \hat{j} - 3 \hat{k})$
Now, A, B, C lie on line L
$A B = (2 - p) \hat{i} - 2 \hat{j} + 2 k$
$B C = - \hat{i} - \hat{j} + \hat{k}$
A, B, C ar collinear
$∴ \frac{2 - p}{- 1} = \frac{- 2}{- 1} = \frac{2}{1}$
$2 - p = - 2 \Rightarrow p = 4$
$d.r. of line (- 1, - 1, 1)$
Equation of plane through part $(- p, p, p + 1)$ i.e., $(- 4, 4, 5)$
$| \begin{array}{ccc} x - 2 & y - 5 & z + 4 \\ - 4 - 2 & 4 - 5 & 5 + 4 \\ - 1 & - 1 & 1 \end{array} | = 0$
$| \begin{array}{ccc} x - 2 & y - 5 & z + 4 \\ - 6 & - 1 & 9 \\ - 1 & - 1 & 1 \end{array} | = 0$
$(x - 2) (- 1 + 9) - (y - 5) (- 6 + 9) + (z + 4) (6 - 1) = 0$
$8 x - 3 y + 3 z = - 19$
$- \frac{8}{19} x + \frac{3}{19} y - \frac{5}{19} z = 1$
$∴ a = - \frac{8}{19}, b = \frac{3}{19}, c = \frac{- 5}{19}$
$a + b + c = - \frac{8}{19} + \frac{3}{19} - \frac{5}{19} = - \frac{10}{19}$
$p (a + b + c) = 4 \times \frac{- 10}{19} = - \frac{40}{19}$

TS EAMCET-2020-11.09.2020

Vector Algebra

88199 Let A, B, C be three points whose position vectors respectively are:
$\vec{a} = \hat{i} + 4 \hat{j} + 3 \hat{k}$
$\vec{b} = 2 \hat{i} + α \hat{j} + 4 \hat{k}, α \in ◻$
$\vec{c} = 3 \hat{i} - 2 \hat{j} + 5 \hat{k}$
If $α$ is the smallest positive integer for which $\vec{a}, \vec{b}, \vec{c}$ are non-collinear, then the length of the median, in $△ ABC$ , through $A$ is :

1

\frac{\sqrt{82}}{2}

2

\frac{\sqrt{62}}{2}

3

\frac{\sqrt{69}}{2}

4

\frac{\sqrt{66}}{2}

Explanation:

(A) : Given that the position vectors-
$\vec{a} = \hat{i} + 4 \hat{j} + 3 \hat{k}$
$\vec{b} = 2 \hat{i} + α \hat{j} + 4 \hat{k}$
$\vec{c} = 3 \hat{i} - 2 \hat{j} + 5 \hat{k}$
Since, $\vec{AB} = \vec{b} - \vec{a}$
$= 2 \hat{i} + α \hat{j} + 4 \hat{k} - (\hat{i} + 4 \hat{j} + 3 \hat{k})$
$= \hat{i} + (α - 4) \hat{j} + \hat{k}$
$\vec{A C} = \vec{c} - \vec{a}$
$= (3 \hat{i} - 2 \hat{j} + 5 \hat{k}) - (\hat{i} + 4 \hat{j} + 3 \hat{k})$
$= 2 \hat{i} - 6 \hat{j} + 2 \hat{k}$
Therefore, $A, B, C$ are collinear, then $\vec{AB}, \vec{AC}$
$\frac{1}{2} = \frac{α - 4}{- 6} = \frac{1}{2}$
$α = 1$
So, is $α = 2$ is the smallest positive integer for whose A, $B, C$ are non collinear
$∴$ Mid-point of $BC = P (\frac{5}{2}, 0, \frac{9}{2})$
Length of the median through $A = A P$
$= \sqrt{{(\frac{5}{2} - 1)}^{2} + 4^{2} + {(\frac{9}{2} - 3)}^{2}}$
$= \sqrt{\frac{9}{4} + 16 + \frac{9}{4}} = \frac{\sqrt{82}}{2}$

JEE Main-2022-29.06.2022

Vector Algebra

88200 If $a, b, c$ are non-coplanar vectors and $λ$ is a real number, then the vectors $a + 2 b + 3 c, λ b +$ $4 c$ and $(2 λ - 1) c$ are non-coplanar for

1 all value of

λ

2 all exactly one values of

λ

3 all exactly two values of

λ

4 no value of

λ

Explanation:

(C) : Given that, the three vectors $(a + 2 b + 3 c)$ , $(λ b + 4 c)$ and $(2 λ - 1) c$
$∴$ Given three vectors are non-coplanar
$| \begin{array}{llc} 1 & 2 & 3 \\ 0 & λ & 4 \\ 0 & 0 & 2 λ - 1 \end{array} | \neq 0$
$λ (2 λ - 1) - 0 \neq 0 \Rightarrow λ \neq 0 and λ \neq \frac{1}{2}$
Hence, $a, b$ and $c$ are non coplanar for all exactly two values of $λ$ .

AIEEE-2004

Vector Algebra

88198 $l, m, n$ are three unit vectors in a right handed system and $L$ is a line through the points $A, B$ , C whose position vectors are $p l + 7 m - 6 n, 2 l +$ $5 m - 4 n$ and $l + 4 m - 3 n$ respectively. If the equation of the plane containing $L$ and the points $(- p, p, p + 1)$ is $a x + by + cz = l$ , then $p (a$ $+ b + c) =$

1 0

2

\frac{- 40}{19}

3

\frac{40}{19}

4 -6

Explanation:

(B) : Let $l, m, n$ be $\hat{i}, \hat{j}, \hat{k}$ $A (p \hat{i} + 7 \hat{j} - 6 \hat{k}), B (2 \hat{i} + 5 \hat{j} - 4 \hat{k})$ and $C (\hat{i} + 4 \hat{j} - 3 \hat{k})$
Now, A, B, C lie on line L
$A B = (2 - p) \hat{i} - 2 \hat{j} + 2 k$
$B C = - \hat{i} - \hat{j} + \hat{k}$
A, B, C ar collinear
$∴ \frac{2 - p}{- 1} = \frac{- 2}{- 1} = \frac{2}{1}$
$2 - p = - 2 \Rightarrow p = 4$
$d.r. of line (- 1, - 1, 1)$
Equation of plane through part $(- p, p, p + 1)$ i.e., $(- 4, 4, 5)$
$| \begin{array}{ccc} x - 2 & y - 5 & z + 4 \\ - 4 - 2 & 4 - 5 & 5 + 4 \\ - 1 & - 1 & 1 \end{array} | = 0$
$| \begin{array}{ccc} x - 2 & y - 5 & z + 4 \\ - 6 & - 1 & 9 \\ - 1 & - 1 & 1 \end{array} | = 0$
$(x - 2) (- 1 + 9) - (y - 5) (- 6 + 9) + (z + 4) (6 - 1) = 0$
$8 x - 3 y + 3 z = - 19$
$- \frac{8}{19} x + \frac{3}{19} y - \frac{5}{19} z = 1$
$∴ a = - \frac{8}{19}, b = \frac{3}{19}, c = \frac{- 5}{19}$
$a + b + c = - \frac{8}{19} + \frac{3}{19} - \frac{5}{19} = - \frac{10}{19}$
$p (a + b + c) = 4 \times \frac{- 10}{19} = - \frac{40}{19}$

TS EAMCET-2020-11.09.2020

Vector Algebra

88199 Let A, B, C be three points whose position vectors respectively are:
$\vec{a} = \hat{i} + 4 \hat{j} + 3 \hat{k}$
$\vec{b} = 2 \hat{i} + α \hat{j} + 4 \hat{k}, α \in ◻$
$\vec{c} = 3 \hat{i} - 2 \hat{j} + 5 \hat{k}$
If $α$ is the smallest positive integer for which $\vec{a}, \vec{b}, \vec{c}$ are non-collinear, then the length of the median, in $△ ABC$ , through $A$ is :

1

\frac{\sqrt{82}}{2}

2

\frac{\sqrt{62}}{2}

3

\frac{\sqrt{69}}{2}

4

\frac{\sqrt{66}}{2}

Explanation:

(A) : Given that the position vectors-
$\vec{a} = \hat{i} + 4 \hat{j} + 3 \hat{k}$
$\vec{b} = 2 \hat{i} + α \hat{j} + 4 \hat{k}$
$\vec{c} = 3 \hat{i} - 2 \hat{j} + 5 \hat{k}$
Since, $\vec{AB} = \vec{b} - \vec{a}$
$= 2 \hat{i} + α \hat{j} + 4 \hat{k} - (\hat{i} + 4 \hat{j} + 3 \hat{k})$
$= \hat{i} + (α - 4) \hat{j} + \hat{k}$
$\vec{A C} = \vec{c} - \vec{a}$
$= (3 \hat{i} - 2 \hat{j} + 5 \hat{k}) - (\hat{i} + 4 \hat{j} + 3 \hat{k})$
$= 2 \hat{i} - 6 \hat{j} + 2 \hat{k}$
Therefore, $A, B, C$ are collinear, then $\vec{AB}, \vec{AC}$
$\frac{1}{2} = \frac{α - 4}{- 6} = \frac{1}{2}$
$α = 1$
So, is $α = 2$ is the smallest positive integer for whose A, $B, C$ are non collinear
$∴$ Mid-point of $BC = P (\frac{5}{2}, 0, \frac{9}{2})$
Length of the median through $A = A P$
$= \sqrt{{(\frac{5}{2} - 1)}^{2} + 4^{2} + {(\frac{9}{2} - 3)}^{2}}$
$= \sqrt{\frac{9}{4} + 16 + \frac{9}{4}} = \frac{\sqrt{82}}{2}$

JEE Main-2022-29.06.2022

Vector Algebra

88200 If $a, b, c$ are non-coplanar vectors and $λ$ is a real number, then the vectors $a + 2 b + 3 c, λ b +$ $4 c$ and $(2 λ - 1) c$ are non-coplanar for

1 all value of

λ

2 all exactly one values of

λ

3 all exactly two values of

λ

4 no value of

λ

Explanation:

(C) : Given that, the three vectors $(a + 2 b + 3 c)$ , $(λ b + 4 c)$ and $(2 λ - 1) c$
$∴$ Given three vectors are non-coplanar
$| \begin{array}{llc} 1 & 2 & 3 \\ 0 & λ & 4 \\ 0 & 0 & 2 λ - 1 \end{array} | \neq 0$
$λ (2 λ - 1) - 0 \neq 0 \Rightarrow λ \neq 0 and λ \neq \frac{1}{2}$
Hence, $a, b$ and $c$ are non coplanar for all exactly two values of $λ$ .

AIEEE-2004

Vector Algebra

88198 $l, m, n$ are three unit vectors in a right handed system and $L$ is a line through the points $A, B$ , C whose position vectors are $p l + 7 m - 6 n, 2 l +$ $5 m - 4 n$ and $l + 4 m - 3 n$ respectively. If the equation of the plane containing $L$ and the points $(- p, p, p + 1)$ is $a x + by + cz = l$ , then $p (a$ $+ b + c) =$

1 0

2

\frac{- 40}{19}

3

\frac{40}{19}

4 -6

Explanation:

(B) : Let $l, m, n$ be $\hat{i}, \hat{j}, \hat{k}$ $A (p \hat{i} + 7 \hat{j} - 6 \hat{k}), B (2 \hat{i} + 5 \hat{j} - 4 \hat{k})$ and $C (\hat{i} + 4 \hat{j} - 3 \hat{k})$
Now, A, B, C lie on line L
$A B = (2 - p) \hat{i} - 2 \hat{j} + 2 k$
$B C = - \hat{i} - \hat{j} + \hat{k}$
A, B, C ar collinear
$∴ \frac{2 - p}{- 1} = \frac{- 2}{- 1} = \frac{2}{1}$
$2 - p = - 2 \Rightarrow p = 4$
$d.r. of line (- 1, - 1, 1)$
Equation of plane through part $(- p, p, p + 1)$ i.e., $(- 4, 4, 5)$
$| \begin{array}{ccc} x - 2 & y - 5 & z + 4 \\ - 4 - 2 & 4 - 5 & 5 + 4 \\ - 1 & - 1 & 1 \end{array} | = 0$
$| \begin{array}{ccc} x - 2 & y - 5 & z + 4 \\ - 6 & - 1 & 9 \\ - 1 & - 1 & 1 \end{array} | = 0$
$(x - 2) (- 1 + 9) - (y - 5) (- 6 + 9) + (z + 4) (6 - 1) = 0$
$8 x - 3 y + 3 z = - 19$
$- \frac{8}{19} x + \frac{3}{19} y - \frac{5}{19} z = 1$
$∴ a = - \frac{8}{19}, b = \frac{3}{19}, c = \frac{- 5}{19}$
$a + b + c = - \frac{8}{19} + \frac{3}{19} - \frac{5}{19} = - \frac{10}{19}$
$p (a + b + c) = 4 \times \frac{- 10}{19} = - \frac{40}{19}$

TS EAMCET-2020-11.09.2020

Vector Algebra

88199 Let A, B, C be three points whose position vectors respectively are:
$\vec{a} = \hat{i} + 4 \hat{j} + 3 \hat{k}$
$\vec{b} = 2 \hat{i} + α \hat{j} + 4 \hat{k}, α \in ◻$
$\vec{c} = 3 \hat{i} - 2 \hat{j} + 5 \hat{k}$
If $α$ is the smallest positive integer for which $\vec{a}, \vec{b}, \vec{c}$ are non-collinear, then the length of the median, in $△ ABC$ , through $A$ is :

1

\frac{\sqrt{82}}{2}

2

\frac{\sqrt{62}}{2}

3

\frac{\sqrt{69}}{2}

4

\frac{\sqrt{66}}{2}

Explanation:

(A) : Given that the position vectors-
$\vec{a} = \hat{i} + 4 \hat{j} + 3 \hat{k}$
$\vec{b} = 2 \hat{i} + α \hat{j} + 4 \hat{k}$
$\vec{c} = 3 \hat{i} - 2 \hat{j} + 5 \hat{k}$
Since, $\vec{AB} = \vec{b} - \vec{a}$
$= 2 \hat{i} + α \hat{j} + 4 \hat{k} - (\hat{i} + 4 \hat{j} + 3 \hat{k})$
$= \hat{i} + (α - 4) \hat{j} + \hat{k}$
$\vec{A C} = \vec{c} - \vec{a}$
$= (3 \hat{i} - 2 \hat{j} + 5 \hat{k}) - (\hat{i} + 4 \hat{j} + 3 \hat{k})$
$= 2 \hat{i} - 6 \hat{j} + 2 \hat{k}$
Therefore, $A, B, C$ are collinear, then $\vec{AB}, \vec{AC}$
$\frac{1}{2} = \frac{α - 4}{- 6} = \frac{1}{2}$
$α = 1$
So, is $α = 2$ is the smallest positive integer for whose A, $B, C$ are non collinear
$∴$ Mid-point of $BC = P (\frac{5}{2}, 0, \frac{9}{2})$
Length of the median through $A = A P$
$= \sqrt{{(\frac{5}{2} - 1)}^{2} + 4^{2} + {(\frac{9}{2} - 3)}^{2}}$
$= \sqrt{\frac{9}{4} + 16 + \frac{9}{4}} = \frac{\sqrt{82}}{2}$

JEE Main-2022-29.06.2022

Vector Algebra

88200 If $a, b, c$ are non-coplanar vectors and $λ$ is a real number, then the vectors $a + 2 b + 3 c, λ b +$ $4 c$ and $(2 λ - 1) c$ are non-coplanar for

1 all value of

λ

2 all exactly one values of

λ

3 all exactly two values of

λ

4 no value of

λ

Explanation:

(C) : Given that, the three vectors $(a + 2 b + 3 c)$ , $(λ b + 4 c)$ and $(2 λ - 1) c$
$∴$ Given three vectors are non-coplanar
$| \begin{array}{llc} 1 & 2 & 3 \\ 0 & λ & 4 \\ 0 & 0 & 2 λ - 1 \end{array} | \neq 0$
$λ (2 λ - 1) - 0 \neq 0 \Rightarrow λ \neq 0 and λ \neq \frac{1}{2}$
Hence, $a, b$ and $c$ are non coplanar for all exactly two values of $λ$ .

AIEEE-2004

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Question Bank Series

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