SATHEE: Vectors 4 Question 1

Vectors 4 Question 1

1. Let $a = \hat{i} - \hat{j}, b = \hat{i} + \hat{j} + \hat{k}$ and $c$ be a vector such that $a \times c + b = 0$ and $a \cdot c = 4$ , then $| c |^{2}$ is equal to

(2019 Main, 9 Jan I)

(a) 8

(b) $\frac{19}{2}$

(d) $\frac{17}{2}$

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Answer:

Correct Answer: 1. (b)

Solution:

We have, $(a \times c) + b = 0$

$\Rightarrow a \times (a \times c) + a \times b = 0$

(taking cross product with a on both sides)

$\begin{aligned} \Rightarrow (a \cdot c) a - (a \cdot a) c + | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 1 & 0 \\ 1 & 1 & 1 \end{array} | = 0 \\ [∵ a \times (b \times c) = (a \cdot c) b - (a \cdot b) c] \\ \Rightarrow 4 (\hat{i} - \hat{j}) - 2 c + (- \hat{i} - \hat{j} + 2 \hat{k}) = 0 \\ [∵ a \cdot a = (\hat{i} - \hat{j}) (\hat{i} - \hat{j}) = 1 + 1 = 2 and a \cdot c = 4] \\ \Rightarrow 2 c = 4 i - 4 \hat{j} - \hat{i} - \hat{j} + 2 \hat{k} \\ \Rightarrow c = \frac{3 \hat{i} - 5 \hat{j} + 2 \hat{k}}{2} \Rightarrow | c |^{2} = \frac{9 + 25 + 4}{4} = \frac{19}{2} \end{aligned}$

Vectors 4 Question 1

1. Let $a = \hat{i} - \hat{j}, b = \hat{i} + \hat{j} + \hat{k}$ and $c$ be a vector such that $a \times c + b = 0$ and $a \cdot c = 4$ , then $| c |^{2}$ is equal to

Answer:

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Vectors 4 Question 1

1. Let a=i^−j^,b=i^+j^+k^ and c be a vector such that a×c+b=0 and a⋅c=4, then |c|2 is equal to

Answer:

Engineering Preparation

Medical Preparation

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Syllabus

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NCERT Exercise Video Solution

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1. Let $a = \hat{i} - \hat{j}, b = \hat{i} + \hat{j} + \hat{k}$ and $c$ be a vector such that $a \times c + b = 0$ and $a \cdot c = 4$ , then $| c |^{2}$ is equal to