SATHEE: Vectors 3 Question 2

Vectors 3 Question 2

2. 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

(2019 Main, 11 Jan I)

(a) $- 10 \hat{i} + 5 \hat{j}$

(b) $- 10 \hat{i} - 5 \hat{j}$

(d) $- 14 \hat{i} + 5 \hat{j}$

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Answer:

Correct Answer: 2. (a)

Solution:

We know that, if $a, b, c$ are coplanar vectors, then $[\begin{array}{lll} a & b & c \end{array}] = 0$

$\begin{aligned} ∴ | \begin{array}{ccc} 1 & 2 & 4 \\ 1 & λ & 4 \\ 2 & 4 & λ^{2} - 1 \end{array} | = 0 \\ \Rightarrow 1 λ (λ^{2} - 1) - 16 - 2 ((λ^{2} - 1) - 8) + 4 (4 - 2 λ) = 0 \\ \Rightarrow λ^{3} - λ - 16 - 2 λ^{2} + 18 + 16 - 8 λ = 0 \\ \Rightarrow λ^{3} - 2 λ^{2} - 9 λ + 18 = 0 \\ \Rightarrow λ^{2} (λ - 2) - 9 (λ - 2) = 0 \end{aligned}$

$\begin{aligned} \Rightarrow (λ - 2) (λ^{2} - 9) = 0 \\ \Rightarrow (λ - 2) (λ + 3) (λ - 3) = 0 \\ ∴ λ = 2, 3 or - 3 \\ If λ = 2, then \\ \begin{array}{r} a \times c = | \begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 4 \\ 2 & 4 & 3 \end{array} | = \hat{i} (6 - 16) - \hat{j} (3 - 8) + \hat{k} (4 - 4) \\ = - 10 \hat{i} + 5 \hat{j} \end{array} \\ If λ = \pm 3, then a \times c = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 4 \\ 2 & 4 & 8 \end{array} | = 0 \end{aligned}$

(because last two rows are proportional).

Vectors 3 Question 2

2. 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

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Vectors 3 Question 2

2. Let a=i^+2j^+4k^,b=i^+λj^+4k^ and c=2i^+4j^+(λ2−1)k^ be coplanar vectors. Then, the non-zero vector a×c is

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2. 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