Let Z, a= i+j-k and b=3 i-j+2 k. Let c be a vector such that (a+b… - Vidyadip

MCQ

Let $\lambda \in Z, \vec{a}=\lambda \hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}$. Let $\overrightarrow{ c }$ be a vector such that $(\vec{a}+\vec{b}+\vec{c}) \times \vec{c}=\overrightarrow{0}, \vec{a} \cdot \vec{c}=-17$ and $\vec{b} \cdot \vec{c}=-20$ Then $|\overrightarrow{ c } \times(\lambda \hat{i}+\hat{j}+\hat{ k })|^2$ is equal to

A
$62$
✓
$46$
C
$53$
D
$49$

✓

Answer

Correct option: B.

$46$

b
$(\vec{a}+\vec{b}+\overrightarrow{ c }) \times \overrightarrow{ c }=0$

$(\vec{a}+\vec{b}) \times \vec{c}=0$

$\overrightarrow{ c }=\alpha(\overrightarrow{ a }+\overrightarrow{ b })=\alpha(\lambda+3) \hat{ i }+\alpha \hat{ k }$

$\vec{b} \cdot \vec{c}=-20 \Rightarrow 3 \alpha(\lambda+3)+2 \alpha=-20$

$\vec{a} . \vec{c}=-17 \Rightarrow \alpha \lambda(\lambda+3)-\alpha=-17$

$\Rightarrow \alpha(3 \lambda+9+2)=-20$

$\alpha\left(\lambda^2+3 \lambda-1\right)=-17$

$17(3 \lambda+11)=20\left(\lambda^2+3 \lambda-1\right)$

$20 \lambda^2+9 \lambda-207=0$

$\lambda=3 \quad(\lambda \in Z)$

$\Rightarrow \alpha=-1 \quad \Rightarrow \overrightarrow{ c }=-(6 \hat{ i }+\hat{ k })$

$\overrightarrow{ v }=\overrightarrow{ c } \times(3 \hat{ i }+\hat{ j }+\hat{ k })$

$=\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ -6 & 0 & -1 \\ 3 & 1 & 1\end{array}\right|=\hat{i}+3 \hat{j}-6 \hat{k}$

$|\overrightarrow{ v }|^2=(-1)^2+3^2+6^2=46$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from 10. vector algebra→10. vector algebra — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Find the direction cosines of the line $\frac{x-2}{2}=\frac{2 y-5}{-3}=\frac{z+1}{0}$.

Let $\text{f(x)=}\begin{cases}\frac{\text{x}-4}{|\text{x}-4|}+\text{a},&\text{if }\text{ x} < 4\\\text{a}+\text{b},&\text{if }\text{ x} =4\\\frac{\text{x}-4}{|\text{x}-4|}+\text{b},&\text{if }\text{ x} > 4\end{cases}$ Then, f(x) is continus at x = 4 when:

a = 0, b = 0
a = 1, b = 1
a = -1, b = 1
a = 1, b = -1

The value of $\int\limits^{\pi}_0\frac{\text{x}\tan\text{x}}{\sec\text{x}+\cos\text{x}}\text{ dx}$ is:

$\frac{\pi^2}{4}$
$\frac{\pi^2}{2}$
$\frac{3\pi^2}{2}$
$\frac{\pi^2}{2}$

Let X denote the number of times heads occur in n tosses of a fair coin. If P(X = 4), P(X = 5) and P(X = 6) are in AP, the value of n is:

Write the direction cosines of a line parallel to the line $\frac{3-x}{3}=\frac{y+2}{-2}=\frac{z+2}{6}$.

$\int_1^2 {{e^x}\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)\,dx = } $

If $f(x) = 2{x^3} - 21{x^2} + 36x - 30$, then which one of the following is correct

$\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{99}}+{{2}^{99}}+{{3}^{99}}+......{{n}^{99}}}{{{n}^{100}}}=$

The points of discontinuity of the function $\text{f(x)}=\begin{cases}\frac{1}{5}(2\text{x}^2+3),&\text{x}\leq1\\6-5\text{x},&1<\text{x}<3\\\text{x}-3,&\text{x}\geq3\end{cases}$ is (are):

x = 1
x = 3
x = 1, 3
none of these

If $f(x)$ is a differentiable function such that $f:R \to r$ and $f\left( {\frac{1}{n}} \right) = 0\;\forall \;n \ge 1,n \in I$ then