Let a = i + 4j, b = 4j + k and c = i - 2k. If d is a vector perpendicular to both a and b such that c * d = 16, then, d, is equal to Option 1: √33 Option 2: 2√33 Option 3: 3√33 Option 4: 4√33

Answer: 4√33. Solution: We need to find a vector d that is perpendicular to both a and b. If a vector is perpendicular to both a and b, it must be parallel to a x b, meaning d = (a x b) for some scalar . Given: a = i + 4j + 0k, b = 0i + 4j + k a x b = vmatrix i & j & k \\ 1 & 4 & 0 \\ 0 & 4 & 1 vmatrix = i(4 * 1 - 0 * 4) - j(1 * 1 - 0 * 0) + k(1 * 4 - 4 * 0) = 4i - j + 4k Since d is parallel to a x b: d = (4i - j + 4k) Using c * d = 16 where c = i + 0j - 2k: c * d = [(1)(4) + (0)(-1) + (-2)(4)] = (4 + 0 - 8) = -4 = 16 = -4 d = -4(4i - j + 4k) = -16i + 4j - 16k, d, = √((-16)^2 + (4)^2 + (-16)^2) = √(256 + 16 + 256) = √528 = √(16 x 33) = 4√33, d, = 4√33

Algebra

Vector Algebra

Medium

core

Let $\vec{a} = \hat{i} + 4\hat{j}$ , $\vec{b} = 4\hat{j} + \hat{k}$ and $\vec{c} = \hat{i} - 2\hat{k}$ . If $\vec{d}$ is a vector perpendicular to both $\vec{a}$ and $\vec{b}$ such that $\vec{c} \cdot \vec{d} = 16$ , then $|\vec{d}|$ is equal to

$\sqrt{33}$

$2\sqrt{33}$

$3\sqrt{33}$

$4\sqrt{33}$

✅ Correct Option: 4

We need to find a vector $\vec{d}$ that is perpendicular to both $\vec{a}$ and $\vec{b}$ . If a vector is perpendicular to both $\vec{a}$ and $\vec{b}$ , it must be parallel to $\vec{a} \times \vec{b}$ , meaning $\vec{d} = \lambda(\vec{a} \times \vec{b})$ for some scalar $\lambda$ .

Given:

$\vec{a} = \hat{i} + 4\hat{j} + 0\hat{k}$ , $\quad \vec{b} = 0\hat{i} + 4\hat{j} + \hat{k}$

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & 0 \\ 0 & 4 & 1 \end{vmatrix}$

$= \hat{i}(4 \cdot 1 - 0 \cdot 4) - \hat{j}(1 \cdot 1 - 0 \cdot 0) + \hat{k}(1 \cdot 4 - 4 \cdot 0)$

$= 4\hat{i} - \hat{j} + 4\hat{k}$

Since $\vec{d}$ is parallel to $\vec{a} \times \vec{b}$ :

$\vec{d} = \lambda(4\hat{i} - \hat{j} + 4\hat{k})$

Using $\vec{c} \cdot \vec{d} = 16$ where $\vec{c} = \hat{i} + 0\hat{j} - 2\hat{k}$ :

$\vec{c} \cdot \vec{d} = \lambda\big[(1)(4) + (0)(-1) + (-2)(4)\big]$

$= \lambda(4 + 0 - 8)$

$= -4\lambda = 16$

$\lambda = -4$

$\vec{d} = -4(4\hat{i} - \hat{j} + 4\hat{k}) = -16\hat{i} + 4\hat{j} - 16\hat{k}$

$|\vec{d}| = \sqrt{(-16)^2 + (4)^2 + (-16)^2}$

$= \sqrt{256 + 16 + 256}$

$= \sqrt{528}$

$= \sqrt{16 \times 33}$

$= 4\sqrt{33}$

$|\vec{d}| = 4\sqrt{33}$

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If $\vec{a}, \vec{b}$ and $\vec{c}$ are three vectors such that $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$ , where $\vec{a}$ and $\vec{b}$ are unit vectors and $|\vec{c}|=2$ , then the angle between the vectors $\vec{b}$ and $\vec{c}$ is :

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If $|\vec{a}| = |\vec{b}| = |\vec{a} + \vec{b}| = 1$ , then $|\vec{a} - \vec{b}|$ is equal to:

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Let $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = -2\hat{i} + 3\hat{j} - 4\hat{k}$ , then which of the following statements are correct?

(A) $|\vec{a}| = \sqrt{14}$

(B) $|\vec{b}| = 29$

(D) Angle between $\vec{a}$ and $\vec{b}$ is $\cos^{-1}\left(\frac{-8}{\sqrt{406}}\right)$

Choose the correct answer from the options given below:

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The vector $\{(a^2+a-2)\hat{i}+(a^2-4)\hat{j}-(a^2-5a+6)\hat{k}\}$ makes acute angles with coordinate axes, then values of $a$ lies in :

2022: 17 Aug Shift 2

Arrange the vectors in descending order of their magnitudes.

(A) $\hat{i} + \hat{j} + \hat{k}$

(B) $2\hat{i} - 3\hat{j}$

(D) $2\hat{i} - \hat{k}$

Choose the correct answer from the options given below :

Solution

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