\(90^\circ\)
To find the angle between the diagonals AC and BD of the quadrilateral ABCD, we need to determine the vectors for AC and BD using the given position vectors of the vertices A, B, C, and D.
- First, let's determine the vector \(\overrightarrow{AC}\). Given:
- Position vector of A: \(3\hat{i} +4\hat{j}-2\hat{k}\)
- Position vector of C: \(2\hat{i} - 3\hat{j} + 2\hat{k}\)
- Now, let's determine the vector \(\overrightarrow{BD}\). Given:
- Position vector of B: \(4\hat{i} - 4\hat{j} - 3\hat{k}\)
- Position vector of D: \(6\hat{i} - 2\hat{j} + \hat{k}\)
- Next, we compute the dot product of vectors \(\overrightarrow{AC}\) and \(\overrightarrow{BD}\):
- Dot product: \((-1)(2) + (-7)(2) + (4)(4) = -2 - 14 + 16 = 0\)
- The vectors are perpendicular if their dot product is zero. Therefore, the angle between \(\overrightarrow{AC}\) and \(\overrightarrow{BD}\) is \(90^\circ\).
Thus, the correct answer is \(90^\circ\).
Similar Questions
Let \(\vec{a}, \vec{b}, \vec{c}\) be three non-zero vectors such that \(\vec{a}\times \vec{b} = \vec{c} \) . Consider the following statements:
1. \(\vec a\) is unique if \(\vec b\) and \(\vec c\) are given
2. \(\vec c\) is unique if \(\vec a\) and \(\vec b\) are given
Which of the above statements is/are correct?
In a right angled triangle ABC, if the hypotenuse AC = p, then what is \(\overrightarrow {{\rm{AB}}} \cdot \overrightarrow {{\rm{AC}}} + \overrightarrow {{\rm{BC}}} \cdot \overrightarrow {{\rm{BA}}} + \overrightarrow {{\rm{CA}}} \cdot \overrightarrow {{\rm{CB}}} \) equal to?
What is \({\rm{\vec c}}\) equal to?
If \({\rm{\vec d}} = {\rm{x\hat i}} + {\rm{y\hat j}} + {\rm{z\hat k}}\) , then which of the following equations is/are correct?
1. y – x = 4
2. 2z – 3 = 0
Select the correct answer using the code given below:
What is \({\rm{\vec a}} \cdot {\rm{\vec b}} + {\rm{\vec b}} \cdot {\rm{\vec c}} + {\rm{\vec c}} \cdot {\rm{\vec a}}\) equal to?
What is the angle between \({\rm{\vec a}}\) and \({\rm{\vec b}}\) ?
- What is the fourth vertex \(D\)?
- Let \(\vec{a} = \hat{i} - \hat{j} + \hat{k}\) and \(\vec{b} = \hat{i} + 2\hat{j} - \hat{k}\). If \(\vec{a} \times (\vec{b} \times \vec{a}) = \alpha\hat{i} - \beta\hat{j} + \gamma\hat{k}\), then what is the value of \(\alpha + \beta + \gamma\)?
- Let \(\vec{p}=\vec{a}-\vec{b}\), \(\vec{q}=\vec{a}+\vec{b}\). If \(|\vec{a}|=|\vec{b}|= 2\) and \(\vec{a}\cdot\vec{b} = 2\), then what is the value of \(|\vec{p}\times\vec{q}|\)?
How many of the following can be a vector perpendicular to both the vectors \(2\hat{i} - \hat{j} + \hat{k}\) and \(\hat{i} + \hat{j} + 3\hat{k}\) ?
I. \(4\hat{i} + 5\hat{j} - 3\hat{k}\)
II. \(-8\hat{i} - 10\hat{j} + 6\hat{k}\)
III. \(\frac{1}{50}(-4\hat{i}-5\hat{j}+3\hat{k})\)
Select the correct answer.
Important Questions from Vector Algebra
Vector A̅ = ŷ.3 + ẑ.2 and B̅ = x̂.5 + ŷ.8 extend from the origin. Find A̅.B̅ Choose the correct answer.
The value of the cross product \(\left( {\overrightarrow a - \overrightarrow b } \right) \times \left( {\overrightarrow a + \overrightarrow b } \right)\) of two vectors \(\overrightarrow a - \overrightarrow b\) and \(\overrightarrow a + \overrightarrow b \) is:
If non - zero a, b, c are such that a + b + c = 0, then the value of \(\frac{a^2}{bc} + \frac{b^2}{ac} + \frac{c^2}{ab}\) is
Vector a = 3i + 2j – 6k, vector b = 4i – 3j + k, angle between above vectors is
Two forces F1 and F2 are used to pull a car, which met an accident. The angle between the two force is θ. Find the value of θ for the resultant force is equal to \(\sqrt{(F_1^2 + F_2^2)}\)
