SATHEE: Vectors 2 Question 1

Vectors 2 Question 1

1. Let $a = 3 \hat{i} + 2 \hat{j} + 2 \hat{k}$ and $b = \hat{i} + 2 \hat{j} - 2 \hat{k}$ be two vectors. If a vector perpendicular to both the vectors $a + b$ and $a - b$ has the magnitude 12 , then one such vector is

(2019 Main, 12 April II)

(a) $4 (2 \hat{i} + 2 \hat{j} + \hat{k})$

(b) $4 (2 \hat{i} - 2 \hat{j} - \hat{k})$

(d) $4 (- 2 \hat{i} - 2 \hat{j} + \hat{k})$

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Answer:

Correct Answer: 1. (b)

Solution:

Given vectors are

$a = 3 \hat{i} + 2 \hat{j} + 2 \hat{k} and b = \hat{i} + 2 \hat{j} - 2 \hat{k}$

Now, vectors $a + b = 4 \hat{i} + 4 \hat{j}$ and $a - b = 2 \hat{i} + 4 \hat{k}$

$∴$ A vector which is perpendicular to both the vectors $a + b$ and $a - b$ is

$\begin{aligned} (a + b) \times & (a - b) = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{array} | \\ = \hat{i} (16) - \hat{j} (16) + \hat{k} (- 8) = 8 (2 \hat{i} - 2 \hat{j} - \hat{k}) \end{aligned}$

Then, the required vector along $(a + b) \times (a - b)$ having magnitude 12 is

$\pm 12 \times \frac{8 (2 \hat{i} - 2 \hat{j} - \hat{k})}{8 \times \sqrt{4 + 4 + 1}} = \pm 4 (2 \hat{i} - 2 \hat{j} - \hat{k})$

Vectors 2 Question 1

1. Let $a = 3 \hat{i} + 2 \hat{j} + 2 \hat{k}$ and $b = \hat{i} + 2 \hat{j} - 2 \hat{k}$ be two vectors. If a vector perpendicular to both the vectors $a + b$ and $a - b$ has the magnitude 12 , then one such vector is

Answer:

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Vectors 2 Question 1

1. Let a=3i^+2j^+2k^ and b=i^+2j^−2k^ be two vectors. If a vector perpendicular to both the vectors a+b and a−b has the magnitude 12 , then one such vector is

Answer:

Engineering Preparation

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Important Resources

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Sample Mock Test

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NCERT Exercise Video Solution

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1. Let $a = 3 \hat{i} + 2 \hat{j} + 2 \hat{k}$ and $b = \hat{i} + 2 \hat{j} - 2 \hat{k}$ be two vectors. If a vector perpendicular to both the vectors $a + b$ and $a - b$ has the magnitude 12 , then one such vector is