SATHEE: Vectors 2 Question 2

Vectors 2 Question 2

2. Let $α \in R$ and the three vectors $a = α \hat{i} + \hat{j} + 3 \hat{k}, b = 2 \hat{i} + \hat{j} - α \hat{k}$ and $c = α \hat{i} - 2 \hat{j} + 3 \hat{k}$ . Then, the set $S = α : a, b and c$ are coplanar

(2019 Main, 12 April II)

(a) is singleton

(b) is empty

(d) contains exactly two numbers only one of which is positive

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Answer:

Correct Answer: 2. (b)

Solution:

Given three vectors are

$\begin{aligned} a = α \hat{i} + \hat{j} + 3 \hat{k} \\ b = 2 \hat{i} + \hat{j} - α \hat{k} \\ and c = α \hat{i} - 2 \hat{j} + 3 \hat{k} \\ Clearly, [a b c] = | \begin{array}{ccc} α & 1 & 3 \\ 2 & 1 & - α \\ α & - 2 & 3 \end{array} | \\ = α (3 - 2 α) - 1 (6 + α^{2}) + 3 (- 4 - α) \\ = - 3 α^{2} - 18 = - 3 (α^{2} + 6) \end{aligned}$

$∵$ There is no value of $α$ for which $- 3 (α^{2} + 6)$ becomes zero, so $= | \begin{array}{ccc} α & 1 & 3 2 & 1 & - α α & - 2 & 3 \end{array} | [a b c] \neq 0$

$\Rightarrow$ vectors $a, b$ and $c$ are not coplanar for any value $α \in R$ .

So, the $set S = α : a, b and c$ are coplanar is empty set.

Vectors 2 Question 2

2. Let $α \in R$ and the three vectors $a = α \hat{i} + \hat{j} + 3 \hat{k}, b = 2 \hat{i} + \hat{j} - α \hat{k}$ and $c = α \hat{i} - 2 \hat{j} + 3 \hat{k}$ . Then, the set $S = α : a, b and c$ are coplanar

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Vectors 2 Question 2

2. Let α∈R and the three vectors a=αi^+j^+3k^,b=2i^+j^−αk^ and c=αi^−2j^+3k^. Then, the set S=α:a,bandc are coplanar

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2. Let $α \in R$ and the three vectors $a = α \hat{i} + \hat{j} + 3 \hat{k}, b = 2 \hat{i} + \hat{j} - α \hat{k}$ and $c = α \hat{i} - 2 \hat{j} + 3 \hat{k}$ . Then, the set $S = α : a, b and c$ are coplanar