SATHEE: Vectors 4 Question 11

Vectors 4 Question 11

11. Consider the cube in the first octant with sides $O P$ , $O Q$ and $O R$ of length 1, along the $X$ -axis, $Y$ -axis and $Z$ -axis, respectively, where $O (0, 0, 0)$ is the origin. Let $S (\frac{1}{2}, \frac{1}{2}, \frac{1}{2})$ be the centre of the cube and $T$ be the vertex of the cube opposite to the origin $O$ such that $S$ lies on the diagonal OT. If $p = S P, q = S Q, r = S R$ and $t = S T$ , then the value of $| (p \times q) \times (r \times t) |$ is ….. .

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Answer:

Correct Answer: 11. (0.5)

Solution:

(0.5) Here, $P (1, 0, 0), Q (0, 1, 0), R (0, 0, 1), T = (1, 1, 1)$ and $S = (\frac{1}{2}, \frac{1}{2}, \frac{1}{2})$ .

Now,

$\vec{p} = \vec{S P} = \vec{O P} - \vec{O S}$

$= (\frac{1}{2} \hat{i} - \frac{1}{2} \hat{j} - \frac{1}{2} \hat{k}) = \frac{1}{2} (\hat{i} - \hat{j} - \hat{k})$

$\vec{q} = \vec{S Q} = \frac{1}{2} (- \hat{i} + \hat{j} - \hat{k})$

$\vec{r} = \vec{S R} = \frac{1}{2} (- \hat{i} - \hat{j} + \hat{k})$

and $\vec{t} = \vec{S T} = \frac{1}{2} (\hat{i} + \hat{j} + \hat{k})$

$\vec{p} \times \vec{q} = \frac{1}{4} | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} 1 & - 1 & - 1 - 1 & 1 & - 1 \end{array} | = \frac{1}{4} (2 \hat{i} + 2 \hat{j})$

and $\vec{r} \times \vec{t} = \frac{1}{4} | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} - 1 & - 1 & 1 1 & 1 & 1 \end{array} | = \frac{1}{4} (- 2 \hat{i} + 2 \hat{j})$

Now, $(\vec{p} \times \vec{q}) \times (\vec{r} \times \vec{t}) = \frac{1}{16} | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} 2 & 2 & 0 - 2 & 2 & 0 \end{array} |$

$= \frac{1}{16} (8 \hat{k}) = \frac{1}{2} \hat{k}$

$∴ | (p \times q) \times (\vec{r} \times \vec{t}) | = | \frac{1}{2} \hat{k} | = \frac{1}{2} = 0.5$

Vectors 4 Question 11

Answer:

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Vectors 4 Question 11

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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