SATHEE: Vectors 5 Question 5

Vectors 5 Question 5

5. A non-zero vector $\vec{a}$ is parallel to the line of intersection

of the plane determined by the vectors $\hat{i}, \hat{i} + \hat{j}$ and the plane determined by the vectors $\hat{i} - \hat{j}, \hat{i} + \hat{k}$ . The angle between $\vec{a}$ and the vector $\hat{i} - 2 \hat{j} + 2 \hat{k}$ is…… .

(1996, 2M)

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Answer:

Correct Answer: 5. $\frac{π}{4}$ or $\frac{3 π}{4}$

Solution:

Equation of the plane containing $\hat{i}$ and $\hat{i} + \hat{j}$ is

$\begin{aligned} [(\vec{r} - \hat{i}) \hat{i} (\hat{i} + \hat{j})] = 0 \\ \Rightarrow (\vec{r} - \hat{i}) \cdot [\hat{i} \times (\hat{i} + \hat{j})] = 0 \\ \Rightarrow (x \hat{i} + y \hat{j} + z \hat{k}) - \hat{i} \cdot [\hat{i} \times \hat{i} + \hat{i} \times \hat{j}] = 0 \\ \Rightarrow (x - 1) \hat{i} + y \hat{j} + z \hat{k} \cdot [\hat{k}] = 0 \\ \Rightarrow (x - 1) \hat{i} \cdot \hat{k} + y \hat{j} \cdot \hat{k} + z \hat{k} \cdot \hat{k} = 0 \\ \Rightarrow z = 0 \end{aligned}$

Equation of the plane containing $\hat{i} - \hat{j}$ and $\hat{i} + \hat{k}$ is

$\begin{aligned} [(\vec{r} - (\hat{i} - \hat{j})) (\hat{i} - \hat{j}) (\hat{i} + \hat{k})] = 0 \\ \Rightarrow (\vec{r} - \hat{i} + \hat{j}) \cdot [(\hat{i} - \hat{j}) \times (\hat{i} + \hat{k})] = 0 \\ \Rightarrow (x \hat{i} + y \hat{j} + z \hat{k}) - (\hat{i} - \hat{j}) \cdot [\hat{i} \times \hat{i} + \hat{i} \times \hat{k} - \hat{j} \times \hat{i} - \hat{j} \times \hat{k}] = 0 \\ \Rightarrow (x - 1) \hat{i} + (y + 1) \hat{j} + z \hat{k}) \cdot [- \hat{j} + \hat{k} - \hat{i}] = 0 \\ \Rightarrow - (x - 1) - (y + 1) + z = 0 \end{aligned}$

Let $\vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$

Since, $\vec{a}$ is parallel to Eqs. (i) and (ii), we obtain

$a_{3} = 0$

and $a_{1} + a_{2} - a_{3} = 0 \Rightarrow a_{1} = - a_{2}, a_{3} = 0$

Thus, a vector in the direction $\vec{a}$ is $\hat{i} - \hat{j}$ .

If $θ$ is the angle between $\vec{a}$ and $\hat{i} - 2 \hat{j} + 2 \hat{k}$ , then

$\begin{aligned} \cos θ & = \pm \frac{(1) (1) + (- 1) (- 2)}{\sqrt{1 + 1} \sqrt{1 + 4 + 4}} = \pm \frac{3}{\sqrt{2} \cdot 3} \\ \Rightarrow \cos θ & = \pm \frac{1}{\sqrt{2}} \Rightarrow θ = \frac{π}{4} or \frac{3 π}{4} \end{aligned}$

Vectors 5 Question 5

5. A non-zero vector $\vec{a}$ is parallel to the line of intersection

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Vectors 5 Question 5

5. A non-zero vector a→ is parallel to the line of intersection

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

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NCERT Exercise Video Solution

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5. A non-zero vector $\vec{a}$ is parallel to the line of intersection