Let a⃗ = 2î + ĵ − 2k̂, b⃗ = î + ĵ and c⃗ = a⃗ × b⃗. Let d⃗ be a vector such that |d⃗ − a⃗| = √11, |c⃗ × d⃗| = 3 and the angle between c⃗ and d⃗ is π/4. Then a⃗

Let a = 2i + j − 2k, b = i + j and c = a × b. If |d − a| = √11, |c × d| = 3 and angle between c and d is π/4, find a · d

Q. Let $$ \vec a = 2\hat i + \hat j - 2\hat k,\quad \vec b = \hat i + \hat j $$ and $$ \vec c = \vec a \times \vec b. $$ Let $\vec d$ be a vector such that $$ |\vec d-\vec a|=\sqrt{11},\quad |\vec c \times \vec d|=3 $$ and the angle between $\vec c$ and $\vec d$ is $\dfrac{\pi}{4}$. Then $\vec a\cdot\vec d$ is equal to

(A) 0

(B) 1

(D) 11

Correct Answer: 0

Explanation

First compute $\vec c=\vec a\times\vec b$:

$$ \vec c= \begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & 1 & -2\\ 1 & 1 & 0 \end{vmatrix} =2\hat i-2\hat j+\hat k $$

$$ |\vec c|=\sqrt{2^2+(-2)^2+1^2}=3 $$

Using $$ |\vec c\times\vec d|=|\vec c||\vec d|\sin\theta $$ with $\theta=\dfrac{\pi}{4}$:

$$ 3=3|\vec d|\cdot\frac{1}{\sqrt2} \Rightarrow |\vec d|=\sqrt2 $$

Now use the identity

$$ |\vec d-\vec a|^2=|\vec d|^2+|\vec a|^2-2\vec a\cdot\vec d $$

Here,

$$ |\vec a|^2=2^2+1^2+(-2)^2=9,\quad |\vec d|^2=2 $$

Given $|\vec d-\vec a|=\sqrt{11}$:

$$ 11=2+9-2\vec a\cdot\vec d $$

$$ 11=11-2\vec a\cdot\vec d $$

$$ \vec a\cdot\vec d=0 $$

Explanation

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