1
JEE Main 2024 (Online) 4th April Evening Shift
+4
-1

For $$\lambda>0$$, let $$\theta$$ be the angle between the vectors $$\vec{a}=\hat{i}+\lambda \hat{j}-3 \hat{k}$$ and $$\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}$$. If the vectors $$\vec{a}+\vec{b}$$ and $$\vec{a}-\vec{b}$$ are mutually perpendicular, then the value of (14 cos $$\theta)^2$$ is equal to

A
25
B
50
C
20
D
40
2
JEE Main 2024 (Online) 4th April Evening Shift
+4
-1

Let $$\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}+4 \hat{j}-5 \hat{k}$$ and $$\vec{c}=x \hat{i}+2 \hat{j}+3 \hat{k}, x \in \mathbb{R}$$. If $$\vec{d}$$ is the unit vector in the direction of $$\vec{b}+\vec{c}$$ such that $$\vec{a} \cdot \vec{d}=1$$, then $$(\vec{a} \times \vec{b}) \cdot \vec{c}$$ is equal to

A
3
B
9
C
11
D
6
3
JEE Main 2024 (Online) 4th April Morning Shift
+4
-1

Let a unit vector which makes an angle of $$60^{\circ}$$ with $$2 \hat{i}+2 \hat{j}-\hat{k}$$ and an angle of $$45^{\circ}$$ with $$\hat{i}-\hat{k}$$ be $$\vec{C}$$. Then $$\vec{C}+\left(-\frac{1}{2} \hat{i}+\frac{1}{3 \sqrt{2}} \hat{j}-\frac{\sqrt{2}}{3} \hat{k}\right)$$ is:

A
$$-\frac{\sqrt{2}}{3} \hat{i}+\frac{\sqrt{2}}{3} \hat{j}+\left(\frac{1}{2}+\frac{2 \sqrt{2}}{3}\right) \hat{k}$$
B
$$\left(\frac{1}{\sqrt{3}}+\frac{1}{2}\right) \hat{i}+\left(\frac{1}{\sqrt{3}}-\frac{1}{3 \sqrt{2}}\right) \hat{j}+\left(\frac{1}{\sqrt{3}}+\frac{\sqrt{2}}{3}\right) \hat{k}$$
C
$$\frac{\sqrt{2}}{3} \hat{i}-\frac{1}{2} \hat{k}$$
D
$$\frac{\sqrt{2}}{3} \hat{i}+\frac{1}{3 \sqrt{2}} \hat{j}-\frac{1}{2} \hat{k}$$
4
JEE Main 2024 (Online) 1st February Morning Shift
+4
-1
Let $\overrightarrow{\mathrm{a}}=-5 \hat{i}+\hat{j}-3 \hat{k}, \overrightarrow{\mathrm{b}}=\hat{i}+2 \hat{j}-4 \hat{k}$ and

$\overrightarrow{\mathrm{c}}=(((\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \times \hat{i}) \times \hat{i}) \times \hat{i}$. Then $\vec{c} \cdot(-\hat{i}+\hat{j}+\hat{k})$ is equal to :
A
-12
B
-10
C
-13
D
-15
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