SATHEE: Vectors 2 Question 7

Vectors 2 Question 7

7. Let $a = 2 \hat{i} + \hat{j} - 2 \hat{k}, b = \hat{i} + \hat{j}$ and $c$ be a vector such that $| c - a | = 3, | (a \times b) \times c | = 3$ and the angle between $c$ and $a \times b$ is $30^{\circ}$ . Then, $a \cdot c$ is equal to

(2017 Main)

(a) $\frac{25}{8}$

(b) 2

(d) $\frac{1}{8}$

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Answer:

Correct Answer: 7. (a, c, d)

Solution:

We have, $a = 2 \hat{i} + \hat{j} - 2 \hat{k}$

$\Rightarrow | a | = \sqrt{4 + 1 + 4} = 3$

$and b = \hat{i} + \hat{j}$

$\Rightarrow | b | = \sqrt{1 + 1} = \sqrt{2}$

Now, $| c - a | = 3 \Rightarrow | c - a |^{2} = 9$

$\Rightarrow (c - a) \cdot (c - a) = 9$

$\Rightarrow | c |^{2} + | a |^{2} - 2 c \cdot a = 9$

Again,

$| (a \times b) \times c | = 3$

$\Rightarrow | a \times b | | c | \sin 30^{\circ} = 3$

$\Rightarrow | c | = \frac{6}{| a \times b |}$

But $a \times b = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} 2 & 1 & - 2 1 & 1 & 0 \end{array} | = 2 \hat{i} - 2 \hat{j} + \hat{k}$

$∴ | c | = \frac{6}{\sqrt{4 + 4 + 1}} = 2$

From Eqs. (i) and (ii), we get

$\begin{array}{rr} (2)^{2} + (3)^{2} - 2 c \cdot a = 9 \\ \Rightarrow & 4 + 9 - 2 c \cdot a = 9 \\ \Rightarrow & c \cdot a = 2 \end{array}$

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Vectors 2 Question 7

7. Let $a = 2 \hat{i} + \hat{j} - 2 \hat{k}, b = \hat{i} + \hat{j}$ and $c$ be a vector such that $| c - a | = 3, | (a \times b) \times c | = 3$ and the angle between $c$ and $a \times b$ is $30^{\circ}$ . Then, $a \cdot c$ is equal to

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Vectors 2 Question 7

7. Let a=2i^+j^−2k^,b=i^+j^ and c be a vector such that |c−a|=3,|(a×b)×c|=3 and the angle between c and a×b is 30∘. Then, a⋅c is equal to

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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7. Let $a = 2 \hat{i} + \hat{j} - 2 \hat{k}, b = \hat{i} + \hat{j}$ and $c$ be a vector such that $| c - a | = 3, | (a \times b) \times c | = 3$ and the angle between $c$ and $a \times b$ is $30^{\circ}$ . Then, $a \cdot c$ is equal to